Vektor dan Operasi Vektor

Lanjutan Materi Operasi Vektor di Ruang (Cross Product): Perkalian Silang Dua Vektor

$\color{blue}\textrm{C. Perkalian Silang Vektor (Pengayaan)}$.

Pada ruang dimensi tiga khususnya pada vektor akan berlaku perkalian silang (cross vektor) adalah perkalian antara dua vektor yang menghasilkan vektor tunggal. Misalkan diketahui  $\vec{u}$  dan  $\vec{v}$  adalah dua vektor sembarang dan keduanya membentuk sudut  $\theta$, maka hasil kali kedua vektor tersebut adalah sebuah vektor baru dengan dinotasiakan sebagai  $\vec{u}\times \vec{v}$. Tentunya sebagai syarat kedua vektor tersebut masing-masing tidak berupa vektor nol.

Jika  $\vec{u}\times \vec{v}=\vec{c}$ , maka

$\begin{aligned}\vec{u}&\times \vec{v}=\vec{c}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k} \end{aligned}$

Lalu kalau sudah demikian berapa besarnya? dan ke mana arahnya?

Besarnya adalah  $\left | \vec{u}\times \vec{v} \right |=\left | \vec{u} \right |\left | \vec{v} \right |\sin \theta$  dan arahnya tegak lurus terhadap  $\vec{u}$  dan  $\vec{v}$.

Sebagai ilustrasi perhatikanlah gambar berikut  untuk dua buah vektor sebagai misal  $\vec{a}$  dan  $\vec{b}$.

Jika putarannya dibalik, maka akan mendapatkan hasil sebagai mana ilustrasi berikut

Sehingga perlu diingat bahwa :  $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$.

Pada hasil kali silang dua vektor berlaku

1. tidak bersifat komutatif , karena  $\vec{a}\times \vec{b}=-\vec{b}\times \vec{a}$.
2. distributif terhadap penjumlahan : $\vec{a}\times \left (\vec{b}+\vec{c} \right )=\vec{a}\times \vec{b}+\vec{a}\times \vec{c}$.
3. pada perkalian dengan skalar : $k\left (\vec{a}\times \vec{b} \right )=\left (k\vec{a} \right )\times \vec{b}=\vec{a}\times \left ( k\vec{b} \right )$.
4. berlaku untuk sembarang vektor : $\vec{a}\times \vec{a} =0$.
5. jika kedua vektor sejajar, maka hasil kalinya adalah = 0.
6. Nilai dari perkalian kedua vektor terbut adalah sama dengan hasil luas jajar genjang.
7. Nilai dari poin 6 jika dibagi 2 akan berupa hasil luas sebuah segitiga yang dibentuk oleh kedua vektor tersebut.
8. berlaku identitas Lagrange : $\left | \vec{a}\times \vec{b} \right |^{2}=\left | \vec{a} \right |^{2}.\left | \vec{b} \right |^{2}-\left ( \vec{a}\: \bullet \: \vec{b} \right )^{2}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui}\: \: \vec{a}=4\vec{i}+3\vec{j}\: \: \textrm{dan}\: \: \vec{b}=4\vec{i}-3\vec{k}\\ &\textrm{Tentukanlah hasil}\: \: \vec{a}\times \vec{b}\: \: \textrm{dan}\: \: \vec{b}\times \vec{a}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\vec{a}&\times \vec{b}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 4 & 3 &0 \\ 4 & 0 &-3 \end{vmatrix}\\ &=(-9-0)\vec{i}-(-12-0)\vec{j}+(0-12)\vec{k}\\ &=-9\vec{i}+12\vec{j}-12\vec{k} \end{aligned}\\ &\begin{aligned}\vec{b}&\times \vec{a}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{2} & y_{2} &z_{2} \\ x_{1} & y_{1} &z_{1} \end{vmatrix}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 4 & 0 &-3 \\ 4 & 3 &0 \end{vmatrix}\\ &=(0-(-9))\vec{i}-(0-(-12))\vec{j}+(12-0)\vec{k}\\ &=9\vec{i}-12\vec{j}+12\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui}\: \: \vec{a}=6\vec{i}+2\vec{j}+10\vec{k}\: \: \textrm{dan}\: \: \vec{b}=4\vec{i}+\vec{j}+9\vec{k}\\ &\textrm{Tentukanlah hasil}\: \: \vec{a}\times \vec{b}\: \: \textrm{dan}\: \: \vec{b}\times \vec{a}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\vec{a}&\times \vec{b}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 6 & 2 &10 \\ 4 & 1 &9 \end{vmatrix}\\ &=(18-10)\vec{i}-(54-40)\vec{j}+(6-8)\vec{k}\\ &=8\vec{i}-14\vec{j}-2\vec{k} \end{aligned}\\ &\begin{aligned}\vec{b}&\times \vec{a}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{2} & y_{2} &z_{2} \\ x_{1} & y_{1} &z_{1} \end{vmatrix}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ 4 & 1 &9 \\ 6 & 2 &10 \end{vmatrix}\\ &=(10-18)\vec{i}-(40-54)\vec{j}+(8-6)\vec{k}\\ &=-8\vec{i}+14\vec{j}+2\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah luas segitiga}\: \: ABC\: \: \textrm{jika}\\ &\textrm{diketahui}\: \: A(2,1,-2),\: B(0,-1,0),\: \: \textrm{dan}\\ &C(-1,2,-1)\\\\ &\textbf{Jawab}:\\ &\textrm{Misalkan luas segitiga}\: \: \displaystyle \frac{1}{2}\left | \vec{p}\times \vec{q} \right |,\: \: \textrm{dengan}\\ &\begin{cases} \vec{p} & =\overline{AB}=\overline{OB}-\overline{OA}=\begin{pmatrix} 0\\ -1\\ 0 \end{pmatrix}-\begin{pmatrix} 2\\ 1\\ -2 \end{pmatrix}=\begin{pmatrix} -2\\ -2\\ 2 \end{pmatrix} \\ \vec{q} & =\overline{AC}=\overline{OC}-\overline{OA}=\begin{pmatrix} -1\\ 2\\ -1 \end{pmatrix}-\begin{pmatrix} 2\\ 1\\ -2 \end{pmatrix}=\begin{pmatrix} -3\\ 1\\ 1 \end{pmatrix} \end{cases}\\ &\begin{aligned}\vec{p}&\times \vec{q}=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ x_{1} & y_{1} &z_{1} \\ x_{2} & y_{2} &z_{2} \end{vmatrix}\\ &=(y_{1}z_{2}-z_{1}y_{2})\vec{i}-(x_{1}z_{2}-z_{1}x_{2})\vec{j}+(x_{1}y_{2}-y_{1}x_{2})\vec{k}\\ &=\begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -2 & -2 &2 \\ -3 & 1 &1 \end{vmatrix}\\ &=(-2-2)\vec{i}-(-2-(-6))\vec{j}+(-2-6)\vec{k}\\ &=-4\vec{i}-4\vec{j}-8\vec{k}\\ &\textrm{Sehingga}\\ &\left | \vec{p}\times \vec{q} \right |=\sqrt{(-4)^{2}+(-4)^{2}+(-8)^{2}}\\ &\quad\qquad =\sqrt{16+16+64}=\sqrt{96}=4\sqrt{6}\\ &\textrm{Maka luas segi tiganya adalah}:\\ &\textrm{luas}\: \triangle ABC=\displaystyle \frac{1}{2}\left | \vec{p}\times \vec{q} \right |=\displaystyle \frac{1}{2}\left ( 4\sqrt{6} \right )=\color{red}2\sqrt{6} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT. TIGA SERANGKAI PUSTAKA MANDIRI

Lanjutan Materi Operasi Vektor di Ruang (Dot Product)

 $\begin{array}{ll}\\ 6.&\textrm{Diketahui}\: \: \overrightarrow{a}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix},\\ & \textrm{jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q},\: \: \textrm{maka tentukanlah}\\ &\textrm{nilai}\: \: t\: \: \textrm{adalah}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Karena}&\: \textrm{kedua vektor tersebut saling }\\ \textrm{tegak l}& \textrm{urus maka}\\ \overrightarrow{a}.\overrightarrow{b}&=0\\ \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}&\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ (-2).4&+1.(-1)+3.t=0\\ -8-1&+3t=0\\ 3t&=9\\ t&=\color{red}3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Tentukanlah nilai}\: \: \overrightarrow{a}.\overrightarrow{b}\: \: \textrm{jika}\\ &\textrm{a}.\quad \left | \overrightarrow{a} \right |=4,\: \left | \overrightarrow{b} \right |=6,\: \: \angle \left ( \overrightarrow{a},\overrightarrow{b} \right )=60^{\circ}\\ &\textrm{b}.\quad \overrightarrow{a}=2\vec{i}+\vec{j}-5\vec{k}\: \: \textrm{dan}\: \: \overrightarrow{b}=2\vec{i}-3\vec{k}\\ &\textrm{c}.\quad \overrightarrow{a}=\begin{pmatrix} 0\\ -1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 4\\ -2\\ 1 \end{pmatrix}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad \overrightarrow{a}.\overrightarrow{b}&=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\ &=4.6.\cos 60^{\circ}\\ &=24.\left ( \displaystyle \frac{1}{2} \right )\\ &=12 \end{aligned}\\ &\textrm{b}.\quad \overrightarrow{a}.\overrightarrow{b}=2.2+1.0+(-5).(-3)=4+15=19\\ &\textrm{c}.\quad \overrightarrow{a}.\overrightarrow{b}=0.4+(-1).(-2)+3.1=0+2+3=5 \end{array}$

$\begin{array}{ll} 8.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=3\\ & \textrm{dan}\: \: \vec{a}\bullet \vec{b}=15\sqrt{3}\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\\\\ &\textbf{Jawab}\\ &\textrm{Dari bentuk}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ \textrm{dipe}&\textrm{roleh bentuk}\\ \cos \theta &=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ \cos \theta &=\displaystyle \frac{15\sqrt{3}}{10.3}=\frac{15}{30}\sqrt{3}=\frac{1}{2}\sqrt{3}\\ \cos \theta&=\cos 30^{\circ}\\ \theta &=\color{red}30^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: \color{red}30^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Tentukanlah besar sudut antara vektor}\\ &\overrightarrow{a}=\begin{pmatrix} -1\\ 1\\ 0\end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}.\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{\sqrt{(-1)^{2}+1^{2}}\sqrt{1^{2}+(-2)^{2}+2^{2}}}\\ &=\displaystyle \frac{-1-2+0}{\sqrt{2}\sqrt{9}}\\ &=-\displaystyle \frac{1}{\sqrt{2}}=-\displaystyle \frac{1}{2}\sqrt{2}\\ &=-\cos 45^{\circ}\\ &=\cos \left ( 180^{\circ}-45^{\circ} \right )\\ \cos \theta &=\cos 135^{\circ}\\ \therefore \: \theta &=\color{red}135^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Diketahui bahwa}\: \: \left |\overrightarrow{a} \right |=\sqrt{6} ,\: \: (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})=0\\ & \textrm{dan}\: \: \overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})=3.\: \textrm{Tentukanlah besar}\\ &\textrm{sudut antara}\: \: \overrightarrow{a}\: \: \textrm{dan}\: \: \overrightarrow{b}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{Perhatikan}&\: \textrm{bahwa}\\ (\overrightarrow{a}-\overrightarrow{b})(\overrightarrow{a}+\overrightarrow{b})&=0\\ \left | \overrightarrow{a} \right |^{2}-\left | \overrightarrow{b} \right |^{2}&=0\\ \left | \overrightarrow{a} \right |^{2}&=\left | \overrightarrow{b} \right |^{2}\quad \Rightarrow \quad \left | \overrightarrow{a} \right |=\overrightarrow{b}=\sqrt{6}\\ \textrm{dan}\quad \overrightarrow{a}(\overrightarrow{a}-\overrightarrow{b})&=3\\ \left | \overrightarrow{a} \right |^{2}-\overrightarrow{a}\overrightarrow{b}&=3\\ 6-\overrightarrow{a}\overrightarrow{b}&=3\\ -\overrightarrow{a}\overrightarrow{b}&=3-6=-3\\ \overrightarrow{a}\overrightarrow{b}&=3\\ \left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta &=3\\ \cos \theta &=\displaystyle \frac{3}{\sqrt{6}\sqrt{6}}=\frac{3}{6}=\frac{1}{2}\\ \cos \theta &=\cos 60^{\circ}\\ \therefore \: \: \theta &=\color{red}60^{\circ} \end{aligned} \end{array}$

$\color{blue}\textrm{Berikut dua contoh untuk sudut tidak istimewa}$.

$\begin{array}{ll} 11.&\textrm{Diketahui}\: \: \vec{a} =\vec{i}+2\vec{j}+2\vec{k},\: \: \textrm{dan}\\ & \vec{b}=3\vec{i}+4\vec{j}\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{a} =\vec{i}+2\vec{j}+2\vec{k}=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{a} \right |=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3\\ \triangleright \quad &\vec{b}=3\vec{i}+4\vec{j}=\begin{pmatrix} 3\\ 4\\ 0 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{b} \right |=\sqrt{3^{2}+4^{2}+0^{2}}=\sqrt{25}=5\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \cos \theta &=\displaystyle \frac{\vec{a}\bullet \vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ \cos \theta &=\displaystyle \frac{\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 3\\ 4\\ 0 \end{pmatrix}}{3.5}=\frac{3+8+0}{15}=\frac{11}{15}\\ \cos \theta&=0,733\\ \theta &=\color{red}\arccos \left ( \displaystyle 0.733 \right )\\ &\quad \textrm{gunakan alat bantu tabel trigonometri}\\ &\quad \textrm{atau kalkulator scientific}\\ &=42,9^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: 42,9^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll} 12.&\textrm{Diketahui}\: \: \vec{p} =(1,2,2),\: \: \textrm{dan}\\ & \vec{q}=(3,-2,6)\: .\: \textrm{Tentukan sudut}\\ &\textrm{yang dibentuk oleh}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{p} =(1,2,2)=\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{9}=3\\ \triangleright \quad &\vec{q}=(3,-2,6)=\begin{pmatrix} 3\\ -2\\ 6 \end{pmatrix}\: \: \textrm{dan}\\ &\left | \vec{q} \right |=\sqrt{3^{2}+(-2)^{2}+6^{2}}=\sqrt{49}=7\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \cos \theta &=\displaystyle \frac{\vec{p}\bullet \vec{q}}{\left | \vec{p} \right |\left | \vec{q} \right |}\\ \cos \theta &=\displaystyle \frac{\begin{pmatrix} 1\\ 2\\ 2 \end{pmatrix}\begin{pmatrix} 3\\ -2\\ 6 \end{pmatrix}}{3.7}=\frac{3-4+12}{21}=\frac{11}{21}\\ \cos \theta&=0,524\\ \theta &=\color{red}\arccos \left ( \displaystyle 0.524 \right )\\ &\quad \textrm{gunakan alat bantu tabel trigonometri}\\ &\quad \textrm{atau kalkulator scientific}\\ &=58,4^{\circ}\\ \textrm{Jadi}&\: \textrm{sudut antara keduanya adalah}\: \: 58,4^{\circ} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 13.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}\: \: \textrm{dan}\: \: \overrightarrow{b}\: \: \textrm{memiliki }\\ &\textrm{panjang masing-masing adalah 2 dan 3}\\ &\textrm{serta}\: \: \angle \left ( \overrightarrow{a},\overrightarrow{b}\right )=60^{\circ}.\: \textrm{Carilah nilai}\\ &\textrm{a}.\quad \left | \overrightarrow{a}+\overrightarrow{b} \right |\\\\ &\textrm{b}.\quad \left | \overrightarrow{a}-\overrightarrow{b} \right |\\ &\textrm{b}\quad \textrm{besar sudut antara}\\ &\qquad \left ( \overrightarrow{a}+\overrightarrow{b} \right )\: \: \textrm{dan}\: \: \left ( \overrightarrow{a}-\overrightarrow{b} \right )\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left | \overrightarrow{a}+\overrightarrow{b} \right |^{2}\\ &=\left ( \overrightarrow{a}+\overrightarrow{b} \right )\left ( \overrightarrow{a}+\overrightarrow{b} \right )\\ &=\overrightarrow{a}\overrightarrow{a}+2\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{b}\\ &=\left | \overrightarrow{a} \right |^{2}\cos 0^{\circ}+2\left |\overrightarrow{a} \right |\left |\overrightarrow{b} \right |\cos 60^{\circ}+\left | \overrightarrow{b} \right |^{2}\cos 0^{\circ}\\ &=2^{2}.1+2.2.3.\displaystyle \frac{1}{2}+3^{2}.1\\ &=4+6+9=19\\ &\textrm{Jadi, nilainya adalah}\: \: \left | \overrightarrow{a}+\overrightarrow{b} \right |=\color{red}\sqrt{19} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left | \overrightarrow{a}-\overrightarrow{b} \right |^{2}\\ &=\left ( \overrightarrow{a}-\overrightarrow{b} \right )\left ( \overrightarrow{a}-\overrightarrow{b} \right )\\ &=\overrightarrow{a}\overrightarrow{a}-2\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{b}\\ &=\left | \overrightarrow{a} \right |^{2}\cos 0^{\circ}-2\left |\overrightarrow{a} \right |\left |\overrightarrow{b} \right |\cos 60^{\circ}+\left | \overrightarrow{b} \right |^{2}\cos 0^{\circ}\\ &=2^{2}.1-2.2.3.\displaystyle \frac{1}{2}+3^{2}.1\\ &=4-6+9=7\\ &\textrm{Jadi, nilainya adalah}\: \: \left | \overrightarrow{a}-\overrightarrow{b} \right |=\color{red}\sqrt{7} \end{aligned}\\ &\begin{aligned}\textrm{c}.\quad \textrm{Untuk menentukan nilai}&\\ \cos \angle \left ( \overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b} \right )&=\displaystyle \frac{\left (\overrightarrow{a}+\overrightarrow{b} \right ).\left (\overrightarrow{a}-\overrightarrow{b} \right )}{\left | \overrightarrow{a}+\overrightarrow{b} \right |.\left | \overrightarrow{a}-\overrightarrow{b} \right |}\\ &=\displaystyle \frac{\overrightarrow{a}\overrightarrow{a}-\overrightarrow{a}\overrightarrow{b}+\overrightarrow{b}\overrightarrow{a}-\overrightarrow{b}\overrightarrow{b}}{\sqrt{19}.\sqrt{7}}\\ &=\displaystyle \frac{2^{2}-3^{2}}{\sqrt{133}}=-\frac{5}{\sqrt{133}}\\ \angle \left ( \overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b} \right )&=\color{red}\arccos \left ( -\frac{5}{\sqrt{133}} \right ) \end{aligned} \end{array}$

$\color{blue}\textrm{Berikut contoh untuk bentuk sudutnya}$.

$\begin{array}{ll} 14.&\textrm{Diketahui}\: \: \vec{p} =(x,3,2),\: \: \textrm{dan}\\ & \vec{q}=(2,-6,3)\: .\: \textrm{Tentukan nilai}\: \: x\\ &\textrm{agar kedua vektor}\\ &\textrm{a}\quad \textrm{membentuk sudut lancip}\\ &\textrm{b}\quad \textrm{membentuk sudut siku-siku}\\ &\textrm{c}\quad \textrm{membentuk sudut tumpul}\\ &\textrm{d}\quad \textrm{sama panjang}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Dik}&\textrm{etahui bahwa}\\ \triangleright \quad &\vec{p} =(x,3,2)=\begin{pmatrix} x\\ 3\\ 2 \end{pmatrix}\: \: \textrm{dan}\\ \triangleright \quad &\vec{q}=(2,-6,3)=\begin{pmatrix} 2\\ -6\\ 3 \end{pmatrix}\\ \end{aligned}\\ &\textrm{Selanjutnya}\\ &\vec{p}\bullet \vec{q}=\begin{pmatrix} x\\ 3\\ 2 \end{pmatrix}\begin{pmatrix} 2\\ -6\\ 3 \end{pmatrix}\\ &\quad =2x-18+6=2x-12\\ &\textrm{Selanjutnya}\\ &\begin{aligned} \textrm{a}\quad&\textbf{Syarat lancip},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}>0\\ &2x-12>0\Leftrightarrow 2x>12\Leftrightarrow x>6\\ \textrm{b}\quad&\textbf{Syarat siku-siku},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}=0\\ &2x-12=0\Leftrightarrow 2x=12\Leftrightarrow x=6\\ \textrm{c}\quad&\textbf{Syarat tumpul},\: \textrm{yaitu}:\: \vec{p}\bullet \vec{q}<0\\ &2x-12<0\Leftrightarrow 2x<12\Leftrightarrow x<6\\ \textrm{d}\quad&\textbf{Syarat panjang kedua vektor sama}\\ & \textrm{yaitu}:\: \left |\vec{p} \right |= \left |\vec{q} \right |,\: \textrm{maka}\\ &\begin{aligned}&\sqrt{x^{2}+3^{2}+2^{2}}=\sqrt{2^{2}+(-6)^{2}+3^{2}}\\ &x^{2}+9+4=4+36+9\\ &x^{2}=36\\ &x=\pm \sqrt{36}=\pm 6\\ &\textrm{Jadi},\: \color{red}x=-6\: \: \color{black}\textrm{atau}\: \: \color{red}x=6 \end{aligned} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
3. Noormandiri, Sucipto, E. 2003. Buku Pelajaran Matematika SMU untuk Kelas 3 Program IPA. Jakarta: ERLANGGA.
4. Yuana, R.A., Indriyastuti. 2017. Perspektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT. TIGA SERANGKAI PUSTAKA MANDIRI.

Operasi Vektor di Ruang (Lanjutan Materi Vektor di Ruang)

Sebelumnya silahkan lihat di sini

 $\color{blue}\textrm{A. Operasi Vektor Dalam Ruang}$

Operasi vektor pada dimensi tiga kurang lebih sama dengan operasi pada vektor berdimensi dua.

$\color{blue}\textrm{A.  Penjumlahan dan Pengurangan}$.

$\begin{aligned}\textrm{Jika}\: & \textrm{diketahui sebagai misal}\\ \bar{u}&=a\bar{i}+b\bar{j}+c\bar{k}\: \: \: \color{red}\textrm{dan}\\ \bar{v}&=p\bar{i}+q\bar{j}+r\bar{k}\\ \textrm{mak}&\textrm{a}\\ \textbf{Pen}&\textbf{jumlahan dua vektor di atas adalah}\\ \bar{u}+\bar{v}&=(a+p)\bar{i}+(b+q)\bar{j}+(c+r)\bar{k}\\ \textbf{dem}&\textbf{ikian juga untuk pengurangan}\\ \bar{u}-\bar{v}&=(a-p)\bar{i}+(b-q)\bar{j}+(c-r)\bar{k}\\ \end{aligned}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Jika diketahui}\: \: \bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\textrm{Tentukanlah hasil dari}\\ &\textrm{a}.\quad \bar{a}+\bar{b}\\ &\textrm{b}.\quad \bar{a}-\bar{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\bar{a}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \bar{b}=\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix},\\ &\textrm{maka}\\ &\bar{a}+\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}+\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1+8\\ 3+(-2)\\ 7+0 \end{pmatrix}=\begin{pmatrix} 9\\ 1\\ 7 \end{pmatrix}\\ &\textrm{Dan untuk}\: \: \bar{a}-\bar{b}\: \: \textrm{adalah}:\\ &\bar{a}-\bar{b}=\begin{pmatrix} 1\\ 3\\ 7 \end{pmatrix}-\begin{pmatrix} 8\\ -2\\ 0 \end{pmatrix}\\ &\qquad =\begin{pmatrix} 1-8\\ 3-(-2)\\ 7-0 \end{pmatrix}=\begin{pmatrix} -7\\ 5\\ 7 \end{pmatrix} \end{aligned} \end{array}$.

$\color{blue}\textrm{B. 1. Perkalian Skalar dengan Vektor}$.

Misalkan suatu skalar   $m$  dan suatu vektor  $\bar{u}=a\bar{i}+b\bar{j}+c\bar{k}$, maka perkalian $m$  dengan vektor  $\bar{u}$ tersebut adalah  $\bar{u}=ma\bar{i}+mb\bar{j}+mc\bar{k}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 2.&\textrm{Jika}\: \: \bar{a}=\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix},\: \: \textrm{tentukanlah nilai}\\ &\textrm{dari}\: \: 2\bar{a}\: \: \: \textrm{dan}\: \: -3\bar{a}\\\\ &\textbf{Jawab}\\ &\begin{aligned}2\bar{a}&=2\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} 4044\\ 4042\\ 4040 \end{pmatrix},\: \: \textrm{dan}\\ -3\bar{a}&=-3\begin{pmatrix} 2022\\ 2021\\ 2020 \end{pmatrix}=\begin{pmatrix} -6066\\ -6063\\ -6060 \end{pmatrix} \end{aligned} \end{array}$

$\color{blue}\textrm{F. 2. Perkalian Skalar Dua Vektor}$.

Hasil dari perkalian skalar dua vektor $\bar{a}$  dan  $\bar{b}$ adalah :  $\bar{a}\: \: \bullet\: \: \bar{b}$.

Dengan

$\bar{a}\: \: \bullet\: \: \bar{b}=\left | \bar{a} \right |\left | \bar{b} \right |\cos \theta$.  sehingga

$\begin{aligned}&\textrm{Tanda dari hasil skalar ini adalah}\\ &\begin{array}{|l|l|l|}\hline \textbf{Besar sudut}\: \: \: \theta &\textbf{Tanda}&\textrm{Bentuk}\\\hline 0^{\circ}\leq \theta < 90^{\circ}&\textrm{Positif}&\color{red}\textrm{Lancip}\\\hline \theta =90^{\circ}&\textrm{Nol}&\textrm{Siku-siku}\\\hline 90^{\circ}< \theta \leq 180^{\circ}&\textrm{Negatif}&\color{blue}\textrm{Tumpul}\\\hline \end{array}\\ &\textrm{Untuk}\: \: \theta \: \: \textrm{berupa sudut istimewa}:\\ &\begin{array}{|c|c|c|c|c|c|c|}\hline \theta &0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}&180^{\circ}\\\hline \cos \theta &1&\displaystyle \frac{1}{2}\sqrt{3}&\displaystyle \frac{1}{2}\sqrt{2}&\displaystyle \frac{1}{2}&0&-1\\\hline \end{array} \end{aligned}$

Adapun secara rumus untuk menentukan besar sudutnya adalah:

$\cos \theta =\displaystyle \frac{\bar{a}\: \: \bullet\: \: \bar{b} }{\left | \bar{a} \right |\left | \bar{b} \right |}$.

Sebagai ilustrasinya perhatikanlah gambar berikut

Selain hasil di atas ada cara lain menyelesaikan perkalian skalar dua vektor, yaitu:

$\begin{aligned}\textrm{Jika}\: & \textrm{diketahui sebagai misal}\\ \bar{u}&=a\bar{i}+b\bar{j}+c\bar{k}\: \: \: \color{red}\textrm{dan}\\ \bar{v}&=p\bar{i}+q\bar{j}+r\bar{k}\\ \textrm{mak}&\textrm{a}\\ \textbf{Per}&\textbf{kalian skalar dua vektor adalah}:\\ \bar{u}\: \bullet \: &\bar{v}=\left ( a\bar{i}+b\bar{j}+c\bar{k} \right )\left ( p\bar{i}+q\bar{j}+r\bar{k} \right )\\ &\: \: =ap.\bar{i}\: \bullet \bar{i}+aq.\bar{i}\: \bullet \: \bar{j}+ar.\bar{i}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +bp.\bar{j}\: \bullet \: \bar{i}+bq.\bar{j}\: \bullet \: \bar{j}+br.\bar{j}\: \bullet \: \bar{k}\\ &\: \: \: \: \: \: \: +cp.\bar{k}\: \bullet \: \bar{i}+cq.\bar{k}\: \bullet \: \bar{j}+cr.\bar{k}\: \bullet \: \bar{k}\\ &\: \: =ap+0+0+0+bq+0+0+0+cr\\ &\: \: =\color{red}a p+b q+c r \end{aligned}$

$\begin{aligned}&\textrm{Sebagai penjelasannya adalah}:\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{i}=\left | \bar{i} \right |\left | \bar{i} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{j}=\left | \bar{i} \right |\left | \bar{j} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{i}\: \: \bullet \: \: \bar{k}=\left | \bar{i} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{j}=0\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{j}=\left | \bar{j} \right |\left | \bar{j} \right |\cos 0^{\circ}=1.1.1=1\\ &\color{red}\triangleright \quad \color{black}\bar{j}\: \: \bullet \: \: \bar{k}=\left | \bar{j} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.0=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{i}=\bar{i}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{j}=\bar{j}\: \: \bullet \: \: \bar{k}=0\\ &\color{red}\triangleright \quad \color{black}\bar{k}\: \: \bullet \: \: \bar{k}=\left | \bar{k} \right |\left | \bar{k} \right |\cos 90^{\circ}=1.1.1=1 \end{aligned}$

$\begin{aligned}&\textrm{Atau jika ditabelkan nilainya}\\ &\begin{array}{|c|c|c|c|}\hline \bar{u}\: \bullet \: \bar{v}&p\bar{i}&q\bar{j}&r\bar{k}\\\hline a\bar{k}&ap&0&0\\ b\bar{j}&0&bq&0\\ c\bar{k}&0&0&cr\\\hline \end{array} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll} 3.&\textrm{Jika}\: \: \vec{a}=\begin{pmatrix} 1\\ 2\\ 4 \end{pmatrix},\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 5\\ 4\\ 0 \end{pmatrix}\\\ & \textrm{tentukanlah nilai}\: \: \textrm{dari}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.5+2.4+4.0=5+8+0=13 \end{aligned} \end{array}$

$\begin{array}{ll} 4.&\textrm{Jika diketahui}\: \: \vec{a}=\vec{i}-2\vec{j}+3\vec{k},\\ & \textrm{dan}\: \: \vec{b}=3\vec{i}-4\vec{j}+m\vec{k}\: \: \textrm{serta}\\\ & \textrm{nilai}\: \: \vec{a}\bullet \vec{b}=-4,\: \: \textrm{maka tentukan}\\ &\textrm{nilai}\: \: m\\\\ &\textbf{Jawab}\\ &\textrm{Diketahui bahwa}\\ &\color{red}\triangleright \quad \color{black}\vec{a}=\vec{i}-2\vec{j}+3\vec{k}=\begin{pmatrix} 1\\ -2\\ 3 \end{pmatrix},\: \: \textrm{dan}\\ &\color{red}\triangleright \quad \color{black}\vec{b}=3\vec{i}-4\vec{j}+m\vec{k}=\begin{pmatrix} 3\\ -4\\ m \end{pmatrix}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=1.3+3.(-4)+3.m\\ -4&=3+8+3m\\ -3m&=11+4\\ m&=-\displaystyle \frac{15}{3}\\ &=\color{blue}-5 \end{aligned} \end{array}$

$\begin{array}{ll} 5.&\textrm{Diketahui}\: \: \left |\vec{a} \right |=10,\: \left | \vec{b} \right |=6.\\ & \textrm{Jika}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{membentuk sudut}\\ &60^{\circ}.\: \textrm{Tentukanlah nilai}\: \: \vec{a}\bullet \vec{b}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\vec{a}\bullet \vec{b}&=\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ &=10.6.\cos 60^{\circ}\\ &=60.\left ( \displaystyle \frac{1}{2} \right )\\ &=\color{blue}30\\ \textrm{Jadi}&\: \textrm{hasil kali skalarnya adalah 30} \end{aligned} \end{array}$.

Vektor pada Ruang

 $\color{blue}\textrm{A. Vektor Di Ruang}$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{|c|c|}\hline \textrm{Nama}&\textbf{R}^{3}\\\hline \textrm{Vektor Satuan}&\textrm{Ruang (Bidang XYZ)}\\\hline \hat{e}_{\bar{a}}=\displaystyle \frac{\bar{a}}{\left | \bar{a} \right |}&\begin{cases} i= &\textrm{vektor satuan} \\ &\textrm{yang searah sumbu X}\\ j= &\textrm{Vektor satuan}\\ &\textrm{yang searah sumbu Y}\\ k=&\textrm{Vektor satuan}\\ &\textrm{searah sumbu Z} \end{cases} \\\hline \textrm{Vektor nol}&\overrightarrow{O}=\begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix}\\\hline \textrm{Vektor posisi}&\overrightarrow{OP}=\vec{p}=\begin{pmatrix} p_{1}\\ p_{2}\\ p_{3} \end{pmatrix}=p_{1}\bar{i}+p_{2}\bar{j}+p_{3}\bar{j}\\\hline \textrm{Besar Vektor}&\overrightarrow{OP}=\sqrt{p_{1}^{2}+p_{2}^{2}+p_{3}^{2}}\\\hline \end{array}$

$\color{blue}\textrm{B. Operasi Vektor}$

$\color{blue}\textrm{1. Sifat-Sifat Aljabar Vektor}$

$\begin{array}{|l|l|l|}\hline 1.&\textrm{Komutatif penjumlahan}&\vec{a}+\vec{b}=\vec{b}+\vec{a}\\\hline 2.&\textrm{Asosiatif penjumlahan}&\left ( \vec{a}+\vec{b} \right )+\vec{c}=\vec{a}+\left ( \vec{b}+\vec{c} \right )\\\hline 3.&\textrm{Elemen Identitas}&\vec{a}+\vec{0}=\vec{0}+\vec{a}=\vec{a}\\\hline 4.&\textrm{Invers Penjumlahan}&\vec{a}+\left ( -\vec{a} \right )=\left ( -\vec{a} \right )+\vec{a}=\vec{0}\\\hline 5.&\textrm{Perkalian dengan skalar}&k\left ( l\vec{a} \right )=\left ( kl \right )\vec{a}\\ &&k\left ( \vec{a}+ \vec{b} \right )=k\vec{a}+k\vec{b}\\ &&k\left ( \vec{a}- \vec{b} \right )=k\vec{a}-k\vec{b}\\\hline 6.&\begin{aligned}&\textrm{Jika A, B, dan C segaris }\\ &\color{blue}\textrm{(Kolinear)} \end{aligned}&\begin{cases} \overrightarrow{AB}=k\overrightarrow{BC} \\ \overrightarrow{AC}=k\overrightarrow{AB} \\ dll \end{cases}\\\hline \end{array}$.

$\begin{array}{|c|c|}\hline \color{blue}\textrm{Vektor}&\color{blue}\textrm{Contoh}\\\hline \vec{z}=a\vec{i}+b\vec{j}+c\vec{k}&\begin{aligned}&\textrm{diketahui}\: \: \vec{p}=\vec{i}-2\vec{j}+2\vec{k}\\ &\textrm{maka pangjang vektor}\: \: \vec{p}\: \: \textrm{adalah}\\ &\left | \vec{p} \right |=\sqrt{1^{2}+(-2)^{2}+2^{2}}\\ &\quad\: \: =\sqrt{1+4+4}=\sqrt{9}=3 \end{aligned}\\\hline &\begin{aligned}&\textrm{Vektor satuan dari}\: \: \vec{p}\: \: \textrm{adalah}\\ &\vec{e}_{\vec{p}}=\frac{\vec{p}}{\left | \vec{p} \right |}=\displaystyle \frac{\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}}{3}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 1\\ -2\\ 2 \end{pmatrix}=\displaystyle \begin{pmatrix} \frac{1}{3}\\ -\frac{2}{3}\\ \frac{2}{3} \end{pmatrix} \end{aligned}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}+\overrightarrow{b}\\ &\textrm{b}.\quad 6\overrightarrow{a}+2\overrightarrow{b}\\ &\textrm{c}.\quad 2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &\textrm{d}.\quad \displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}+\overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 2+(-3)\\ 1+(-5)\\ (-4)+2 \end{pmatrix}\\ &=\begin{pmatrix} 2-3\\ 1-5\\ -4+2 \end{pmatrix}=\color{red}\begin{pmatrix} -1\\ -4\\ -2 \end{pmatrix}\\ \textrm{b}.\quad&6\overrightarrow{a}+2\overrightarrow{b}=6\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}+2\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ 6-10\\ -24+4 \end{pmatrix}=\color{red}\begin{pmatrix} 12\\ -4\\ -20 \end{pmatrix}\\ \textrm{c}.\quad&2\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}\\ &2\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}-\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}+\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=\begin{pmatrix} 4+3+7\\ 2+5+0\\ -8-2+4 \end{pmatrix}=\color{red}\begin{pmatrix} 17\\ 7\\ -6 \end{pmatrix}\\ \textrm{d}.\quad&\displaystyle \frac{1}{2}\overrightarrow{c}-\overrightarrow{a}+\displaystyle \frac{3}{4}\overrightarrow{b}=\cdots \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\textrm{tentukanlah}\: \: \left | \overrightarrow{a} \right |\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\left | \overrightarrow{a} \right |&=\sqrt{2^{2}+1^{2}+(-4)^{2}}\\ &=\sqrt{4+1+16}\\ &=\color{red}\sqrt{21} \end{aligned} \end{array}$

$\color{blue}\textrm{2. Perkalian Skalar Dua Vektor}$

Konsep perkalian skalar dua buah vektor di ruang sama persis dengan konsep di bidang, yaitu:

$\color{red}\overrightarrow{a}\cdot \overrightarrow{b}=\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |\cos \theta$.

Misalkan diketahui

$\color{red}\begin{aligned}&\overrightarrow{a}=\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}, \overrightarrow{b}=\begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix},\: \: \color{black}\textrm{maka}\\ & \overrightarrow{a} \cdot \overrightarrow{b} =\color{black}\begin{pmatrix} a_{1}\\ a_{2}\\ a_{3} \end{pmatrix}\cdot \begin{pmatrix} b_{1}\\ b_{2}\\ b_{3} \end{pmatrix}\\ &\qquad\quad =\color{black}a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3} \end{aligned}$

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor-vektor}\: \overrightarrow{a}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\\ &\overrightarrow{b}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix},\: \: \textrm{dan}\: \: \overrightarrow{c}=\begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix},\\ & \textrm{tentukanlah hasil dari}\\ &\textrm{a}.\quad \overrightarrow{a}\cdot \overrightarrow{b}\\ &\textrm{b}.\quad \overrightarrow{a}\cdot \overrightarrow{c}\\ &\textrm{c}.\quad \overrightarrow{b}\cdot \overrightarrow{c}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}\quad&\overrightarrow{a}\cdot \overrightarrow{b}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\cdot \begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\\ &=(2)(-3)+(1)(-5)+(-4)(2)\\ &=-6-5-8=\color{red}-19\\ \textrm{b}\quad&\overrightarrow{a}\cdot \overrightarrow{c}=\begin{pmatrix} 2\\ 1\\ -4 \end{pmatrix}\cdot \begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=(2)(7)+(1)(0)+(-4)(4)\\ &=14+0-16=\color{red}-2\\ \textrm{c}\quad&\overrightarrow{b}\cdot \overrightarrow{c}=\begin{pmatrix} -3\\ -5\\ 2 \end{pmatrix}\cdot \begin{pmatrix} 7\\ 0\\ 4 \end{pmatrix}\\ &=(-3)(7)+(-5)(0)+(2)(4)\\ &=-21+0+8=\color{red}-13 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah nilai}\: \: t\: \: \textrm{jika}\\ & \overrightarrow{p}=3\bar{i}+t\bar{j}+\bar{k}\: \: \textrm{dan}\: \: \overrightarrow{p}\cdot \overrightarrow{p}=13\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\overrightarrow{p}\cdot \overrightarrow{p}=13\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |\left | \overrightarrow{p} \right |\cos 0^{\circ}=13,\\ &\qquad\qquad \color{blue}\textrm{ingat bahwa}\: \: \angle \left ( \overrightarrow{p},\overrightarrow{p} \right )=0^{\circ}\\ &\qquad\qquad \color{blue}\textrm{dan nilai}\: \: \cos 0^{\circ}=1,\\ & \color{black}\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{p}=\left | \overrightarrow{p} \right |^{2}.1=13\Leftrightarrow \left | \overrightarrow{p} \right |^{2}=13\\ &\Leftrightarrow \left (\sqrt{3^{2}+t^{2}+1^{2}} \right )^{2}=13\\ &\Leftrightarrow 3^{2}+t^{2}+1^{2}=13\\ &\Leftrightarrow 9+t^{2}+1=13\\ &\Leftrightarrow t^{2}=13-9-1=10\\ &\Leftrightarrow t^{2}=3\\ &\Leftrightarrow t=\color{red}\pm \sqrt{3} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui}\: \: \overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{Jika}\: \: \overrightarrow{p}\: \: \textrm{tegak lurus}\: \: \overrightarrow{q}\: \: \textrm{maka}\\ &\textrm{tentukanlah nilai}\: \: t\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\textrm{Diketahui bahwa}\\ &\overrightarrow{p}=\begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{q}=\begin{pmatrix} 4\\ -1\\ t \end{pmatrix}\\ &\textrm{dengan}\: \: \overrightarrow{p}\: \: \textrm{dan}\: \: \overrightarrow{q}\: \: \textrm{tegak lurus}\\ &\textrm{artinya}\: \: \color{blue}\angle \left ( \overrightarrow{p},\overrightarrow{q} \right )=90^{\circ}.\: \color{black}\textrm{Sehingga}\\ &\textrm{nilai}\: \: \color{blue}\cos 90^{\circ}=0\\ &\textrm{maka}\\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cos \theta \\ &\overrightarrow{p}\cdot \overrightarrow{q}=\left | \overrightarrow{p} \right |\left | \overrightarrow{q} \right |\cdot 0=0\\ &\Leftrightarrow \: \begin{pmatrix} -2\\ 1\\ 3 \end{pmatrix}\cdot \begin{pmatrix} 4\\ -1\\ t \end{pmatrix}=0\\ &\Leftrightarrow \: (-2)(4)+(1)(-1)+(3)(t)=0\\ &\Leftrightarrow \: -8-1+3t=0\\ &\Leftrightarrow \: 3t=9\\ &\Leftrightarrow \: t=\color{red}3 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \: \left | \overline{u} \right |=6\: ,\: \left | \overline{v} \right |=4\sqrt{3},\: \: \textrm{dan}\: \: \left | \overline{u}-\overline{v} \right |=8\\ &\textrm{tentukanlah nilai dari}\\ &\textrm{a}.\quad \overline{u}\bullet \overline{v}\\ &\textrm{b}.\quad \left | \overline{u}+\overline{v} \right |\\ &\textrm{c}.\quad \textbf{cosinus}\:  \: \textrm{sudut antara}\: \: \overline{u}\: \: \textrm{dan}\: \: \overline{v}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\: \cdots \\ &\: 2.\overline{u}\bullet \overline{v}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}-\left | \overline{u}-\overline{v} \right |^{2}\\ &\: 2.\overline{u}\bullet \overline{v}=6^{2}+(4\sqrt{3})^{2}-8^{2}\\ &\: 2. \overline{u}\bullet \overline{v}=36+48-64=84-64=20\\ &\quad \overline{u}\bullet \overline{v}=\displaystyle \frac{20}{2}=\color{red}10\\ \textrm{b}.\quad &\left | \overline{u}+\overline{v} \right |^{2}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}+2.\overline{u}\bullet \overline{v}\\ &\left | \overline{u}+\overline{v} \right |^{2}=6^{2}+(4\sqrt{3})^{2}+20\\ &\: \: \quad\qquad =84+20=104\\ &\left | \overline{u}+\overline{v} \right |=\color{red}\sqrt{104}\\ \textrm{c}.\quad &\cos \angle (\overline{u},\, \overline{v})=\displaystyle \frac{\overline{u}\bullet \overline{v}}{\left |\overline{u}  \right |.\left | \overline{v} \right |}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &\quad\qquad\qquad =\displaystyle \frac{10\sqrt{3}}{72}=\color{red}\frac{5}{36}\sqrt{3}\\ &\color{blue}\textbf{Berikut ilustrasi gambarnya}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 5.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

RANGKUMAN 

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DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA. 
3. Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Contoh 7 Soal dan Pembahasan Materi Vektor

$\begin{array}{ll}\\ 31.&\textrm{Diketahui}\: \: \left | \vec{a} \right |=\sqrt{3}\: ,\left | \vec{b} \right |=1,\: \: \textrm{dan}\: \: \left | \vec{a}-\vec{b} \right |=1\\ &\textrm{maka panjang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{3}&&&\textrm{d}.&2\sqrt{2}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\color{red}\sqrt{7}&\textrm{e}.&3 \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\: \, \textrm{sebagaimana pada soal}\\ \left | \vec{a}-\vec{b} \right |^{2}&=\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}-2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta \\ 1^{2}&=\left ( \sqrt{3} \right )^{2}+1^{2}-2.\sqrt{3}.1.\cos \theta \\ 2\sqrt{3}\cos \theta &=3\\ \textrm{maka pan}&\textrm{jang vektor}\: \: \vec{a}+\vec{b}\: \: \textrm{adalah}\\ \left | \vec{a}+\vec{b} \right |&=\sqrt{\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+2\left | \vec{a} \right |\left | \vec{b} \right |\cos \theta}\\ &=\sqrt{\left ( \sqrt{3} \right )^{2}+1^{2}+3}\\ &=\sqrt{3+1+3}\\ &=\color{red}\sqrt{7} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 32.&\textrm{Jika}\: \: \left | \overline{u} \right |=6\: ,\: \left | \overline{v} \right |=4\sqrt{3},\: \: \textrm{dan}\: \: \left | \overline{u}-\overline{v} \right |=8\\ &\textrm{tentukanlah nilai dari}\\ &\textrm{a}.\quad \overline{u}\bullet \overline{v}\\ &\textrm{b}.\quad \left | \overline{u}+\overline{v} \right |\\ &\textrm{c}.\quad \textbf{cosinus}\:  \: \textrm{sudut antara}\: \: \overline{u}\: \: \textrm{dan}\: \: \overline{v}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\: \cdots \\ &\: 2.\overline{u}\bullet \overline{v}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}-\left | \overline{u}-\overline{v} \right |^{2}\\ &\: 2.\overline{u}\bullet \overline{v}=6^{2}+(4\sqrt{3})^{2}-8^{2}\\ &\: 2. \overline{u}\bullet \overline{v}=36+48-64=84-64=20\\ &\quad \overline{u}\bullet \overline{v}=\displaystyle \frac{20}{2}=\color{red}10\\ \textrm{b}.\quad &\left | \overline{u}+\overline{v} \right |^{2}=\left | \overline{u} \right |^{2}+\left | \overline{v} \right |^{2}+2.\overline{u}\bullet \overline{v}\\ &\left | \overline{u}+\overline{v} \right |^{2}=6^{2}+(4\sqrt{3})^{2}+20\\ &\: \: \quad\qquad =84+20=104\\ &\left | \overline{u}+\overline{v} \right |=\color{red}\sqrt{104}\\ \textrm{c}.\quad &\cos \angle (\overline{u},\, \overline{v})=\displaystyle \frac{\overline{u}\bullet \overline{v}}{\left |\overline{u}  \right |.\left | \overline{v} \right |}=\frac{10}{6.(4\sqrt{3})}\times \frac{\sqrt{3}}{\sqrt{3}}\\ &\quad\qquad\qquad =\displaystyle \frac{10\sqrt{3}}{72}=\color{red}\frac{5}{36}\sqrt{3}\\ &\color{blue}\textbf{Berikut ilustrasi gambarnya}  \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 33.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 2\\ -5 \end{pmatrix},\: \vec{q}=\begin{pmatrix} 4\\ 3 \end{pmatrix},\: \textrm{maka }\\ &\textrm{proyeksi skalar ortogonal vektor}\: \vec{p}\\ &\textrm{pada}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{7}{5}\\\\ &\textrm{c}.\quad \displaystyle \frac{8}{5}\\\\ &\textrm{d}.\quad \displaystyle \frac{9}{5}\\\\ &\textrm{e}.\quad \displaystyle 2\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left | \vec{r} \right |&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2\\ -5 \end{pmatrix}.\begin{pmatrix} 4\\ 3 \end{pmatrix}}{\sqrt{4^{2}+3^{2}}}\\ &=\displaystyle \frac{8+(-15)}{\sqrt{25}}\\ &=\left | \displaystyle \frac{-7}{5} \right |\\ &=\color{red}\displaystyle \frac{7}{5} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 34.&\textrm{Panjang Proyeksi vektor}\: \: \vec{a}=\begin{pmatrix} 5\\ 1 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} 0\\ 4 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -1\\ &\textrm{b}.\quad -\displaystyle \frac{1}{2}\\ &\textrm{c}.\quad \color{red}1\\ &\textrm{d}.\quad \displaystyle 2\\ &\textrm{e}.\quad 4\\\\ &\textrm{Jawab}\\ &\begin{aligned}\left |\vec{c} \right |&=\left |\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 5\\ 1 \end{pmatrix}\bullet \begin{pmatrix} 0\\ -4 \end{pmatrix}}{\left | \sqrt{0^{2}+(-4)^{2}} \right |} \right |\\ &=\left |\displaystyle \frac{0-4}{4} \right |=\left |-1 \right |=\color{red}1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 35.&\textrm{Proyeksi vektor ortogonal}\: \: \vec{a}=\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ & \textrm{pada}\: \: \vec{b}=\begin{pmatrix} -1\\ 2 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 2\\ -2 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -2\\ 4 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{c}&=\left ( \displaystyle \frac{\vec{a}\bullet \vec{b}}{\left |\vec{b} \right |^{2}} \right ).\vec{b}\\ &=\left (\displaystyle \frac{\begin{pmatrix} 2\\ -4 \end{pmatrix}\bullet \begin{pmatrix} -1\\ 2 \end{pmatrix}}{(-1)^{2}+2^{2}} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\left (\displaystyle \frac{-2-8}{1+4} \right ).\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=-2\begin{pmatrix} -1\\ 2 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 2\\ -4 \end{pmatrix} \end{aligned} \end{array}$

DAFTAR PUSTAKA

1. Johanes, Kastolan, Sulasim. 2006. Kompetensi Matematika Program IPA 3A SMA Kelas XII Semester Pertama. Jakarta: YUDHISTIRA.
2. Kusnandar, Muharman, I., Indrianti, M. 2017. Pendalaman Buku Teks Matematika SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.
3. Miyanto, Aksin, N., Suparno. 2021. Buku Interaktif Matematika untuk SMA/MA Peminatan Matematika dan Ilmu-Ilmu Alam Kelas X Semester 2. Yogyakarta: INTAN PARIWARA. 
4. Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
5. Yuana, R.A., Indriyastuti. 2017. Persektif Matematika untuk Kelas X SMA dan MA Kelompok Peminatan Matematika dan Ilmu Alam. Solo: PT TIGA SERANGKAI MANDIRI.

Contoh 6 Soal dan Pembahasan Materi Vektor

 $\begin{array}{ll}\\ 26.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} -2\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 8\\ 4 \end{pmatrix},\: \: \textrm{maka}\\ &\textrm{sudut yang dibentuk vektor}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\\ &\textrm{adalah}....\\ &\textrm{a}.\quad 0^{\circ}\\ &\textrm{b}.\quad 60^{\circ}\\ &\textrm{c}.\quad 45^{\circ}\\ &\textrm{d}.\quad 60^{\circ}\\ &\textrm{e}.\quad \color{red}90^{\circ}\\\\ &\textrm{Jawab}\\ & \begin{aligned}\vec{p}.\vec{q}&=\displaystyle \begin{vmatrix} \vec{p} \end{vmatrix}.\begin{vmatrix} \vec{q} \end{vmatrix}.\cos \angle \left (\vec{p},\, \vec{q} \right )\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\displaystyle \frac{\vec{p}.\vec{q}}{\left | \vec{p} \right |.\left | \vec{q} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -2\\ 1 \end{pmatrix}.\begin{pmatrix} 8\\ 4 \end{pmatrix}}{\sqrt{(-2)^{2}+1^{2}}.\sqrt{8^{2}+4^{2}}}\\ &=\displaystyle \frac{-16+16}{\sqrt{20}.\sqrt{80}}\\ &=\displaystyle \frac{0}{40}\\ &=0\\ \cos \angle \left (\vec{p},\, \vec{q} \right )&=\cos 90^{\circ}\\ \angle \left (\vec{p},\, \vec{q} \right )&=\color{red}90^{\circ} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 27.&\textrm{Jika}\: \: \overline{OA}=\begin{pmatrix} 1\\ 2 \end{pmatrix},\: \overline{OB}=\begin{pmatrix} 4\\ 2 \end{pmatrix},\: \textrm{dan}\\ &\theta =\angle \left ( \overline{OA},\: \overline{OB} \right ),\: \textrm{maka}\: \: \tan \theta =....\\\\ &\textrm{a}.\quad \displaystyle \frac{3}{5}\\\\ &\textrm{b}.\quad \displaystyle \frac{9}{16}\\\\ &\textrm{c}.\quad \color{red}\displaystyle \frac{3}{4}\\\\ &\textrm{d}.\quad \displaystyle \frac{4}{3}\\\\ &\textrm{e}.\quad \displaystyle \frac{16}{9}\\\\ &\textrm{Jawab}\\ &\begin{array}{|c|c|}\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 1\\ 2 \end{pmatrix}.\begin{pmatrix} 4\\ 2 \end{pmatrix}}{\sqrt{1^{2}+2^{2}}\sqrt{4^{2}+2^{2}}}\\ &=\displaystyle \frac{4+4}{\sqrt{5}.\sqrt{20}}\\ &=\displaystyle \frac{8}{10} \end{aligned}&\begin{aligned}\sin \theta &=\sqrt{1-\cos ^{2}\theta }\\ &=\sqrt{1-\left ( \displaystyle \frac{8}{10} \right )^{2}}\\ &=\sqrt{\displaystyle \frac{36}{100}}\\ &=\displaystyle \frac{6}{10}\\ & \end{aligned}\\\hline \end{array} \\ &\textrm{Selanjutnya}\\ &\begin{aligned}\tan \theta &=\displaystyle \frac{\sin \theta }{\cos \theta }\\ &=\displaystyle \frac{\displaystyle \frac{6}{10}}{\displaystyle \frac{8}{10}}\\ &=\color{red}\displaystyle \frac{3}{4} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 28.&\textrm{Jika}\: \: \vec{a},\: \vec{b}\: \: \textrm{dan}\: \: \vec{c}\: \: \textrm{adalah vektor satuan dengan}\\ & \vec{a}+\vec{b}+\vec{c}=0.\: \textrm{Nilai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle -3\\ &\textrm{b}.\quad \color{red}\displaystyle -\frac{3}{2}\\ &\textrm{c}.\quad \displaystyle 0\\ &\textrm{d}.\quad \displaystyle \frac{3}{2}\\ &\textrm{e}.\quad \displaystyle 3\\\\ &\textrm{Jawab}\\ & \begin{aligned}\textrm{Karena}&\left\{\begin{matrix} \vec{a},\vec{b},\vec{c}\: \: \textrm{adalah vektor satuan, dan}\\ \vec{a}+\vec{b}+\vec{c}=0\qquad\qquad\qquad\qquad . \end{matrix}\right.\\ \textrm{segitig}&\textrm{a ABC adalah segitiga sama sisi}\\ \vec{a}.\vec{b}=&\left | \vec{a} \right |\left | \vec{b} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{a}.\vec{c}=&\left | \vec{a} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \vec{b}.\vec{c}=&\left | \vec{b} \right |\left | \vec{c} \right |\cos 120^{0}=1.1.\left ( -\frac{1}{2} \right )=-\frac{1}{2}\\ \textrm{Jadi, ni}&\textrm{lai dari}\: \: \vec{a}.\vec{b}+\vec{a}.\vec{c}+\vec{b}.\vec{c}\\ &=\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )+\left ( -\frac{1}{2} \right )=\color{red}-\frac{3}{2} \end{aligned} \end{array}$

$.\: \qquad \color{blue}\textrm{berikut ilustrasinya}$

$\begin{array}{ll}\\ 29.&\textrm{Jika}\: \: \angle \left ( \vec{a},\vec{b} \right )=60^{\circ},\: \: \left |\vec{a} \right |=4\: \: \textrm{dan}\\ &\left |\vec{b} \right |=3,\: \: \textrm{maka}\: \: \vec{a}(\vec{a}-\vec{b})\: \: \textrm{adalah}....\\ &\textrm{a}.\quad 2\\ &\textrm{b}.\quad 4\\ &\textrm{c}.\quad 6\\ &\textrm{d}.\quad 8\\ &\textrm{e}.\quad \color{red}10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{a}(\vec{a}-\vec{b})&=\vec{a}.\vec{a}-\vec{a}.\vec{b}\\ &=\left | \vec{a} \right |\left | \vec{a} \right |\cos 0^{\circ}-\left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}\\ &=\left | \vec{a} \right |^{2}-\left | \vec{a} \right |\left | \vec{b} \right |.\displaystyle \frac{1}{2}\\ &=4^{2}-4.3.\displaystyle \frac{1}{2}\\ &=16-6\\ &=\color{red}10 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 30.&\textrm{Tentukan}\: \: \overline{u}\bullet \overline{v}\: ,\: \textrm{jika diketahui}\\ &\textrm{a}.\quad \left |\overline{u}  \right |=10,\:  \left |\overline{v}  \right |=8\sqrt{3},\: \: \cos \angle (\overline{v},\, \overline{u})=\displaystyle \frac{2}{5}\sqrt{3}\\ &\textrm{b}.\quad \left |\overline{u}  \right |=6\sqrt{3},\:  \left |\overline{v}  \right |=4\sqrt{2},\: \: \cos (\overline{v},\, \overline{u})=30^{\circ}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad &\overline{u}\bullet \overline{v}=\left | \overline{u} \right |.\left | \overline{v} \right |.\cos \angle (\overline{v},\, \overline{u})\\ &\overline{u}\bullet \overline{v}=10.(8\sqrt{3}).\displaystyle \frac{2}{5}\sqrt{3}=2.8.2.3=\color{red}96\\ \textrm{b}.\quad &\overline{u}\bullet \overline{v}=\left | \overline{u} \right |.\left | \overline{v} \right |.\cos \angle (\overline{v},\, \overline{u})\\ &\overline{u}\bullet \overline{v}=(6\sqrt{3}).(4\sqrt{2}).\displaystyle \cos 30^{\circ}\\ &\qquad =(6\sqrt{3}).(4\sqrt{2}).\displaystyle \frac{1}{2}\sqrt{3}\\ &\qquad=\displaystyle \frac{6.4.3.\sqrt{2}}{2} =\color{red}36\sqrt{2}\\   \end{aligned} \end{array}$.

Contoh 5 Soal dan Pembahasan Materi Vektor

  $\begin{array}{ll}\\ 21.&\textrm{Jika}\: \: \vec{g}=\begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{h}=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{g}=\vec{h}\: \: \: \textrm{nilai dari}\: \: 4x-3y=.... \\ &\textrm{a}.\quad \color{red}-5\\ &\textrm{b}.\quad -1\\ &\textrm{c}.\quad 0\\ &\textrm{d}.\quad 5\\ &\textrm{e}.\quad 10\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{g}=\vec{h}\\ \begin{pmatrix} 3^{x+y}\\ 5 \end{pmatrix}&=\begin{pmatrix} 81\\ \displaystyle \frac{y+7}{2} \end{pmatrix}\\ 3^{x+y}&=81=3^{4}\Leftrightarrow x+y=4\\ \displaystyle \frac{y+7}{2}&=5\Leftrightarrow y=10-7=3,\quad \textrm{sehingga}\\ x+y&=4\Leftrightarrow x+3=4\Leftrightarrow x=4-3=1,\\ &\textrm{maka}\\ 4x-3y&=4(1)-3(3)\\ &=4-9=\color{red}-5 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 22.&\textrm{Vektor}\: \: \vec{m}=\begin{pmatrix} -2\\ 5 \end{pmatrix}\: \: \textrm{searah }\\ &\textrm{dengan vektor}\: .... \\ &\textrm{a}.\quad \displaystyle \begin{pmatrix} 2\\ -5 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \begin{pmatrix} 2\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \color{red}\displaystyle \begin{pmatrix} -6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle \begin{pmatrix} -4\\ 5 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \begin{pmatrix} -3\\ 10 \end{pmatrix} \\\\ &\textrm{Jawab}\\ &\begin{aligned}&\textrm{Vektor}\quad \vec{m}\: \: \: \textrm{searah dengan vektor}\: \: k.\vec{m}\\ &\color{blue}k.\vec{m}=k\begin{pmatrix} -2\\ 5 \end{pmatrix}, \color{black}\textrm{dengan}\: \: k\: \: \textrm{skalar positif}\\ &\begin{array}{|c|c|c|}\hline \textrm{a}&\textrm{b}\\\hline \begin{pmatrix} 2\\ -5 \end{pmatrix}=-\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} 2\\ 5 \end{pmatrix}=...\\\hline \textrm{c}&\textrm{d}\\\hline \color{red}\begin{pmatrix} -6\\ 15 \end{pmatrix}=3\begin{pmatrix} -2\\ 5 \end{pmatrix}&\begin{pmatrix} -4\\ 5 \end{pmatrix}=...\\\hline \textrm{e}&\\\hline \begin{pmatrix} -3\\ 10 \end{pmatrix}=...&\\\hline \end{array} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 23.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ &\overline{AD}:\overline{DC}=1:2,\: \: \textrm{maka vektor}\\ & \overline{BD}\: \: \textrm{bila dinyatakan}\\ &\textrm{dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ \textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ \textrm{c}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ \textrm{d}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ \textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$.\: \quad \color{blue}\textrm{Gambar berikut untuk soal 24}$

$\begin{array}{ll}\\ 24.&\textrm{Jika vektor}\: \: \overline{AC}=\vec{p},\: \overline{BC}=\vec{q}\: \: \textrm{dan}\\ & \overline{AD}:\overline{DC}=1:2,\: \textrm{maka vektor}\: \: \overline{BD}\\ &\textrm{bila dinyatakan dalam}\: \: \vec{p}\: \: \textrm{dan}\: \: \vec{q}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \displaystyle \frac{1}{3}\left ( 3\vec{p}-2\vec{q} \right )\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right )\\ &\textrm{c}.\quad \displaystyle \left ( \vec{p}-\displaystyle \frac{1}{3}\vec{q} \right )\\ &\textrm{d}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-2\vec{q} \right )\\ &\textrm{e}.\quad \displaystyle \frac{1}{3}\left ( \vec{p}-\vec{q} \right )\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AC}:\overline{CD}&=3:-2\\ \overline{CD}&=-\displaystyle \frac{2}{3}\, \overline{AC}\\ &=\displaystyle -\frac{2}{3}\vec{p}\\ \textrm{maka}\, ,\qquad&\\ \overline{BD}&=\overline{BC}+\overline{CD}\\ &=\vec{q}+\left ( -\displaystyle \frac{2}{3}\vec{p} \right )\\ &=\color{red}\displaystyle \frac{1}{3}\left ( 3\vec{q}-2\vec{p} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 25.&\textrm{Jika titik}\: \: A(2,6)\: \: \textrm{dan}\: \: B(5,3)\: \: \textrm{demikian }\\ &\textrm{juga titik}\: \: P\: \: \textrm{terletak pada}\: \: \overline{AB}\\ &\textrm{dengan}\: \: \overline{AP}:\overline{PB}=2:1,\: \textrm{maka vektor }\\ &\textrm{posisi}\: \: \vec{p}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 4\\ 5 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} -4\\ 4 \end{pmatrix}\\ &\textrm{d}.\quad \begin{pmatrix} 4\\ 2 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} -4\\ 6 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{AP}:\overline{PB}&=2:1\\ \overline{AP}&=2\, \overline{PB}\\ \vec{p}-\vec{a}&=2\left ( \vec{b}-\vec{p} \right )\\ \vec{p}+2\vec{p}&=\vec{a}+2\vec{b}\\ 3\vec{p}&=\vec{a}+2\vec{b}\\ \vec{p}&=\displaystyle \frac{1}{3}\left ( \vec{a}+2\vec{b} \right )\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 2+2.5\\ 6+2.3 \end{pmatrix}\\ &=\displaystyle \frac{1}{3}\begin{pmatrix} 12\\ 12 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 4\\ 4 \end{pmatrix} \end{aligned} \end{array}$

Contoh 4 Soal dan Pembahasan Materi Vektor

 $\begin{array}{ll}\\ 16.&\textrm{Diketahui titik A(-1,1,0) dan titik B(1,-2,2)}\\ &\textrm{maka panjang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\sqrt{2}&&&\textrm{d}.&\color{red}\sqrt{17}\\ \textrm{b}.&\sqrt{5}&\textrm{c}.&\sqrt{9}&\textrm{e}.&\sqrt{21} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}&\textrm{Diketahui}\: \, \textrm{sebagaimana pada soal}\\ &\textrm{maka pan}\textrm{jang vektor}\: \: \overrightarrow{BA}\: \: \textrm{adalah}\\ &\left | \overrightarrow{BA} \right |=\sqrt{(1-(-1))^{2}+(-2-1)^{2}+(2-0)^{2}}\\ &=\sqrt{2^{2}+(-3)^{2}+2^{2}}\\ &=\sqrt{4+9+4}\\ &=\color{red}\sqrt{17} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 17.&\textrm{Vektor satuan untuk vektor}\: \: \vec{a}=\begin{pmatrix} 2, & 1, &-2 \end{pmatrix}=\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -\displaystyle \frac{2}{3}, & -\displaystyle \frac{1}{3}, &\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{b}.&\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix}&\\\\ \textrm{c}.&\begin{pmatrix} \displaystyle \frac{2}{4}, & \displaystyle \frac{1}{4}, &-\displaystyle \frac{2}{4} \end{pmatrix}&\\\\ \textrm{d}.&\begin{pmatrix} -\displaystyle \frac{2}{4}, & -\displaystyle \frac{1}{4}, &\displaystyle \frac{2}{4} \end{pmatrix}\\\\ \textrm{e}.&\begin{pmatrix} -\displaystyle \frac{2}{9}, &-\displaystyle \frac{1}{9}, & \displaystyle \frac{2}{9} \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor }&\: \textrm{satuan dari vektor}\: \: \vec{a}\: \: \textrm{adalah}:\\ \hat{a}&=\displaystyle \frac{\vec{a}}{\left | \vec{a} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{2^{2}+1^{2}+(-2)^{2}}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{\sqrt{9}}\\ &=\displaystyle \frac{\begin{pmatrix} 2, & 1, & -2 \end{pmatrix}}{3}\\ &=\color{red}\begin{pmatrix} \displaystyle \frac{2}{3}, & \displaystyle \frac{1}{3}, &-\displaystyle \frac{2}{3} \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 18.&\textrm{Jika titik A(-2,3,5) dan B(4,1,-3)},\\ & \textrm{maka vektor posisi AB adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\begin{pmatrix} -6, & 2, &8 \end{pmatrix}&\\ \textrm{b}.&\begin{pmatrix} 8, & 2, &-6 \end{pmatrix}&\\ \textrm{c}.&\color{red}\begin{pmatrix} 6, & -2, &-8 \end{pmatrix}&\\ \textrm{d}.&\begin{pmatrix} -8 & -2, &6 \end{pmatrix}\\ \textrm{e}.&\begin{pmatrix} 2, &4, & 2 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \textrm{dari}\: \: \overrightarrow{AB}\: \: \textrm{adalah}:\\ \overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA}\\ &=\begin{pmatrix} 4\\ 1\\ -3 \end{pmatrix}-\begin{pmatrix} -2\\ 3\\ 5 \end{pmatrix}\\ &=\begin{pmatrix} 4+2\\ 1-3\\ -3-5 \end{pmatrix}\\ &=\begin{pmatrix} 6\\ -2\\ -8 \end{pmatrix}\quad\: \textbf{atau}\\ &=\color{red}\begin{pmatrix} 6, & -2, & -8 \end{pmatrix} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 19.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} 5\\ 8 \end{pmatrix}\\ &\textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: x.y=.... \\ &\textrm{a}.\quad 6\\ &\textrm{b}.\quad \color{red}12\\ &\textrm{c}.\quad 18\\ &\textrm{d}.\quad 24\\ &\textrm{e}.\quad 30\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} ^{2}\log 8x\\ \left ( ^{2}\log x \right )^{y} \end{pmatrix}&=\begin{pmatrix} 5\\ 8 \end{pmatrix},\quad\: \: \textrm{maka}\\ 8x&=2^{5}=32\\ \Leftrightarrow x&=\displaystyle \frac{32}{8}=4\\ \left (^{2}\log 4 \right )^{y}&=8\\ \Leftrightarrow 2^{y}&=8=2^{3}\\ \Leftrightarrow y&=3\\ \textrm{Sehingga}&\\ x.y&=4\times 3=\color{red}12 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 20.&\textrm{Jika}\: \: \vec{p}=\begin{pmatrix} 3x\\ 4x+y \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ & \textrm{sehingga}\: \: \vec{p}=\vec{q}\: \: \: \textrm{nilai dari}\: \: 2x+y=.... \\ &\textrm{a}.\quad -12\\ &\textrm{b}.\quad 0\\ &\textrm{c}.\quad 8\\ &\textrm{d}.\quad \color{red}9\\ &\textrm{e}.\quad 19\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}\: &\: \textrm{bahwa}:\quad \vec{p}=\vec{q}\\ \begin{pmatrix} 3x\\ 4x+y \end{pmatrix}&=\begin{pmatrix} \displaystyle \frac{2x-4}{2}\\ 6 \end{pmatrix}\\ 3x&=\displaystyle \frac{2x-4}{2}\Leftrightarrow 6x=2x-4\\ \Leftrightarrow x&=-1\\ 4(-1)+y&=6\Leftrightarrow -4+y=6\\ \Leftrightarrow y&=6+4\\ y&=10\\ x+y&=(-1)+10=\color{red}9 \end{aligned} \end{array}$

Contoh 3 Soal dan Pembahasan Materi Vektor

 $\begin{array}{ll}\\ 11.&\textrm{Jika vektor}\: \: \vec{a}=\begin{pmatrix} 6\\ -4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{b}=\begin{pmatrix} 3\\ 2 \end{pmatrix},\\ &\textrm{maka}\: \: 3\vec{a}-2\vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix}&&\textrm{d}.\quad \begin{pmatrix} 24\\ 16 \end{pmatrix}\\ \textrm{b}.\quad \begin{pmatrix} 24\\ -16 \end{pmatrix}&\textrm{c}.\quad \begin{pmatrix} 12\\ 16 \end{pmatrix}&\textrm{e}.\quad \begin{pmatrix} -12\\ -16 \end{pmatrix} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}3\vec{a}-2\vec{b}&=3\begin{pmatrix} 6\\ -4 \end{pmatrix}-2\begin{pmatrix} 3\\ 2 \end{pmatrix}\\ &=\begin{pmatrix} 18-6\\ -12-4 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 12\\ -16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 12.&\textrm{Diketahui jajar genjang ABCD }\\ &\textrm{dengan titik E adalah perpotongan }\\ &\textrm{diagonal jajar genjang}. \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{Jika}\: \: \overline{AB}=\vec{b}\: \: \textrm{dan}\: \: \overline{AD}=\vec{a},\: \textrm{maka}\: \: \overline{CE}\\ & \textrm{bila dinyatakan dalam}\: \: \vec{a}\: \: \textrm{dan}\: \: \vec{b}\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right )&\\ \textrm{b}.\quad \displaystyle \frac{1}{2}\left ( \vec{a}-\vec{b} \right )&\\ \textrm{c}.\quad \displaystyle \frac{1}{2}\left ( \vec{b}-\vec{a} \right )&\\ \textrm{d}.\quad \color{red}\displaystyle -\frac{1}{2}\left ( \vec{a}+\vec{b} \right )\\ \textrm{e}.\quad -\displaystyle \frac{1}{2}\left ( 2\vec{a}+\vec{b} \right ) \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned} \overline{AC}&=\overline{AD}+\overline{DC}\\ \overline{CA}&=\overline{CD}+\overline{DA}\\ \overline{CE}&=\displaystyle \frac{1}{2}\, \overline{CA}\\ &=\displaystyle \frac{1}{2}\left ( -\vec{b}-\vec{a} \right )\\ &=\color{red}-\displaystyle \frac{1}{2}\left ( \vec{a}+\vec{b} \right ) \end{aligned} \end{array}$

$\begin{array}{ll}\\ 13.&\textrm{Pada segi enam beraturan ABCDEF},\\ & \textrm{jika}\: \: \overrightarrow{AB}=\vec{u}\: \: \textrm{dan}\: \: \overrightarrow{AF}=\vec{v}\: \: \textrm{maka vektor}\\ &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}=....\\ &\begin{array}{llllllll}\\ \textrm{a}.&2\vec{u}+2\vec{v}&&&\textrm{d}.&\color{red}6\vec{u}+6\vec{v}\\ \textrm{b}.&4\vec{u}+4\vec{v}&\textrm{c}.&5\vec{u}+5\vec{v}&\textrm{e}.&8\vec{u}+8\vec{v} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{aligned}.\: \, \qquad &\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}\\ &=\overrightarrow{AB}+\left (\overrightarrow{AO}+\overrightarrow{OC} \right )+2\overrightarrow{AO}+\left (\overrightarrow{AO}+\overrightarrow{OE} \right )+\overrightarrow{AF}\\ &=\vec{u}+\left (2\vec{u}+\vec{v} \right )+2\left ( \vec{v}+\vec{u} \right )+\left ( 2\vec{v}+\vec{u} \right )+\vec{v}\\ &=\color{red}6\vec{u}+6\vec{v} \end{aligned}$

$\begin{array}{ll}\\ 14.&\textrm{Perhatikanlah juga ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{w}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&\color{red}8\vec{i}-6\vec{j}-13\vec{k}&\\ \textrm{b}.&8\vec{i}-13\vec{j}-6\vec{k}\\ \textrm{c}.&6\vec{i}-8\vec{j}-13\vec{k}&\\ \textrm{d}.&-6\vec{i}+8\vec{j}-13\vec{k}\\ \textrm{e}.&-6\vec{i}-13\vec{j}+8\vec{k} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan juga ilustrasi }\\ &\textrm{gambarnya semisal dengan soal No.1}\\ &\textrm{Misalkan titiknya adalah titik }\\ &\textrm{W dengan koordinat (8,-6,-13)},\\ &\textrm{maka vektor posisi titik }\\ &\textrm{W tersebut adalah}\: \: \overrightarrow{OW}=\vec{w}\\ & \textrm{di mana}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{w}\: \textrm{jika dinyatakan }\\ &\textrm{dalam kombinasi linear adalah}\\ &\vec{w}=\color{red}8\vec{i}-6\vec{j}-13\vec{k} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\textrm{Jika titik Z(4,-5,2)},\: \textrm{maka panjang }\\ &\textrm{vektor posisi titik Z adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&1&&&\textrm{d}.&5\sqrt{2}\\ \textrm{b}.&2\sqrt{5}&\textrm{c}.&\color{red}3\sqrt{5}&\textrm{e}.&5\sqrt{3} \end{array}\\\\ &\textbf{Jawab}\\ &\begin{aligned}\textrm{Vektor posisi}&\: \, \textrm{titik Z tersebut adalah}\\ \overrightarrow{OZ}=\vec{z}&=\begin{pmatrix} 4, & -5, & 2 \end{pmatrix},\\ \textrm{Dan panjang}&\: \, \textrm{vektor}\: \: \vec{z}\: \: \textrm{ini adalah}\\ \left | \vec{z} \right |&=\sqrt{4^{2}+(-5)^{2}+2^{2}}\\ &=\sqrt{16+25+4}\\ &=\sqrt{45}=\sqrt{9\times 5}\\ &=\color{red}3\sqrt{5} \end{aligned} \end{array}$.

Contoh 2 Soal dan Pembahasan Materi Vektor

 $\begin{array}{ll}\\ 6.&\textrm{Diketahui titik}\: \: P(n,2),\: Q(1,-2),\: n>0\\ &\textrm{dan panjang}\: \: \overline{PQ}=5,\: \: \textrm{maka nilai}\: \: n\\ & \textrm{adalah}....\\ &\textrm{a}.\quad 1\\ &\textrm{b}.\quad 2\\ &\textrm{c}.\quad 3\\ &\textrm{d}.\quad \color{red}4\\ &\textrm{e}.\quad 5\\\\ &\textrm{Jawab}\\ &\begin{aligned}\overline{PQ}&=5\\ \sqrt{(x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}}&=5\\ (x_{Q}-x_{P})^{2}+(y_{Q}-y_{P})^{2}&=25\\ (1-n)^{2}+(-2-2)^{2}&=25\\ (1-n)^{2}+16&=25\\ (1-n)^{2}-3^{2}&=0\\ (1+3-n)(1-3-n)&=0\\ \color{red}n=4\: \: \color{black}\textrm{atau}\: \: \color{red}n=-2& \end{aligned} \end{array}$

$\begin{array}{ll}\\ 7.&\textrm{Diketahui vektor}\: \: \vec{u}=\begin{pmatrix} 3\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{v}=\begin{pmatrix} 2\\ -1 \end{pmatrix}.\\ & \textrm{Nilai}\: \: \left | \vec{u}+\vec{v} \right |\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \sqrt{28}\\ &\textrm{b}.\quad \sqrt{30}\\ &\textrm{c}.\quad \color{red}\sqrt{34}\\ &\textrm{d}.\quad \sqrt{44}\\ &\textrm{e}.\quad \sqrt{50}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{u}+\vec{v}&=\begin{pmatrix} 3\\ 4 \end{pmatrix} + \begin{pmatrix} 2\\ -1 \end{pmatrix}\\ &=\begin{pmatrix} 3+2\\ 4+(-1) \end{pmatrix}\\ &=\begin{pmatrix} 5\\ 3 \end{pmatrix}\\ \left | \vec{u}+\vec{v} \right |&=\sqrt{5^{2}+3^{2}}\\ &=\sqrt{25+9}\\ &=\color{red}\sqrt{34} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Vektor satuan}\: \: \vec{u}=\begin{pmatrix} -5\\ -12 \end{pmatrix}\: \: \textrm{adalah}.... \\ &\textrm{a}.\quad \color{red}\displaystyle -\frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{b}.\quad \displaystyle \frac{1}{15}\begin{pmatrix} -5\\ -12 \end{pmatrix}\\ &\textrm{c}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{d}.\quad \displaystyle -\frac{1}{17}\begin{pmatrix} 5\\ 12 \end{pmatrix}\\ &\textrm{e}.\quad \displaystyle \frac{1}{2} \begin{pmatrix} 5\\ 12 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{e}_{\vec{u}}&=\displaystyle \frac{\vec{u}}{\left | \vec{u} \right |},\quad \textrm{maka}\\ \vec{e}_{\begin{pmatrix} -5\\ -12 \end{pmatrix}}&=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\left | \begin{pmatrix} -5\\ -12 \end{pmatrix} \right |}\\ &=\displaystyle \frac{\begin{pmatrix} -5\\ -12 \end{pmatrix}}{\sqrt{(-5)^{2}+(-12)^{2}}}\\ &=\displaystyle \frac{-\begin{pmatrix} 5\\ 12 \end{pmatrix}}{\sqrt{169}}\\ &=\color{red}-\displaystyle \frac{1}{13}\begin{pmatrix} 5\\ 12 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 9.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} 8\\ 7 \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 14 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} 6\\ 13 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 6\\ 15 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ 48 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} 8\\ 7 \end{pmatrix}+\begin{pmatrix} -3\\ 9 \end{pmatrix}\\ &=\begin{pmatrix} 8-3\\ 7+9 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} 5\\ 16 \end{pmatrix} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 10.&\textrm{Jika vektor}\: \: \vec{p}=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}\: \: \textrm{dan}\: \: \vec{q}=\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ & \textrm{Hasil dari}\: \: \vec{p}+\vec{q}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \begin{pmatrix} 6\\ 17 \end{pmatrix}\\ &\textrm{b}.\quad \begin{pmatrix} -6\\ -17 \end{pmatrix}\\ &\textrm{c}.\quad \begin{pmatrix} 4\\ 17 \end{pmatrix}\\ &\textrm{d}.\quad \color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix}\\ &\textrm{e}.\quad \begin{pmatrix} 5\\ -16 \end{pmatrix}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}+\vec{q}&=\begin{pmatrix} ^{2}\log 32\\ -\, ^{3}\log \displaystyle \frac{1}{81} \end{pmatrix}+\begin{pmatrix} -9\\ -21 \end{pmatrix}\\ &=\begin{pmatrix} 5-9\\ 4-21 \end{pmatrix}\\ &=\color{red}\begin{pmatrix} -4\\ -17 \end{pmatrix} \end{aligned} \end{array}$

Contoh 1 Soal dan Pembahasan Materi Vektor

 $\begin{array}{ll}\\ 1.&\textrm{Perhatikanlah ilustrasi gambar berikut} \end{array}$

$\begin{array}{ll}\\ .\, \quad&\textrm{maka vektor}\: \: \vec{u}\: \: \textrm{adalah}\: ....\\ &\begin{array}{llllllll}\\ \textrm{a}.&3\vec{i}+5\vec{j}&&&\textrm{d}.&-3\vec{i}-5\vec{j}\\ \textrm{b}.&5\vec{i}+3\vec{j}&\textrm{c}.&\color{red}-3\vec{i}+5\vec{j}&\textrm{e}.&-5\vec{i}+3\vec{j} \end{array}\\\\ &\textbf{Jawab}\\ &\textrm{Kita perhatikan lagi gambarnya}\\ &\begin{aligned}&\textrm{Vektor}\: \: \vec{u}\: \: \textrm{jika dinyatakan sebagai }\\ &\textrm{kombinasi linear adalah}\: \: \vec{u}=\color{red}-3\vec{i}+5\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 2.&\textrm{Panjang vektor}\: \: \vec{p}=\begin{pmatrix} 4\\ -8 \end{pmatrix}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \sqrt{4}\\ &\textrm{b}.\quad \sqrt{12}\\ &\textrm{b}.\quad \sqrt{20}\\ &\textrm{d}.\quad \color{red}\displaystyle \sqrt{80}\\ &\textrm{e}.\quad \sqrt{100}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{p}&=\begin{pmatrix} 4\\ -8 \end{pmatrix},\: \: \textrm{maka besar dari vektor}\\ \vec{p}&\: \: \textrm{adalah}=\left | \vec{p} \right |=\sqrt{x^{2}+y^{2}}\\ &\textrm{Yaitu}\\ \left | \vec{p} \right |&=\sqrt{4^{2}+(-8)^{2}}\\ &=\sqrt{16+64}=\color{red}\sqrt{80} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 3.&\textrm{Perhatikanlah gambar berikut}\\ & \end{array}$.

$\begin{array}{ll}\\ .\quad &\textrm{Panjang vektor}\: \: \vec{h}\: \: \textrm{tersebut di atas adalah}....\\ &\textrm{a}.\quad 5\\ &\textrm{b}.\quad 7\\ &\textrm{c}.\quad \color{red}10\\ &\textrm{d}.\quad 12\\ &\textrm{e}.\quad 15\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Diketahui}&\quad \vec{h}=\overline{OH}=8\vec{i}+6\vec{j}\\ \vec{h}&=\sqrt{x_{H}^{2}+y_{H}^{2}}\\ &=\sqrt{8^{2}+6^{2}}\qquad \textbf{(ingat tripel Pythagoras)}\\ &=\sqrt{10^{2}}=\color{red}10 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 4.&\textrm{Vektor satuan dari}\: \: \vec{q}=3\vec{i}-4\vec{j}\: \: \textrm{adalah}....\\ &\textrm{a}.\quad \displaystyle \frac{4}{5}\vec{i}-\frac{3}{5}\vec{j}\\ &\textrm{b}.\quad \color{red}\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j}\\ &\textrm{c}.\quad 3\vec{i}-4\vec{j}\\ &\textrm{d}.\quad 4\vec{i}-3\vec{j}\\ &\textrm{e}.\quad 15\vec{i}-20\vec{j}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\vec{q}&=3\vec{i}-4\vec{j}\\ &\textrm{Vektor satuan dari vektor}\: \: \vec{q}\: \: \textrm{adalah}:\\ &\: \hat{e}_{_{\vec{q}}}=\displaystyle \frac{1}{\left | \vec{q} \right |}. \vec{q}\\ &\textrm{Sehingga}\\ \hat{e}_{_{\vec{q}}}&=\displaystyle \frac{1}{\sqrt{3^{2}+(-4)^{2}}}.\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &=\displaystyle \frac{1}{5}\begin{pmatrix} 3\\ -4 \end{pmatrix}\\ &\textrm{atau dalam vektor basis}\\ &=\color{red}\displaystyle \frac{3}{5}\vec{i}-\frac{4}{5}\vec{j} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Vektor berikut yang memiliki panjang}\\ & 29\: \: \textrm{satuan adalah}....\\ &\textrm{a}.\quad \displaystyle 18\vec{i}-19\vec{j}\\ &\textrm{b}.\quad \displaystyle 19\vec{i}-20\vec{j}\\ &\textrm{c}.\quad 20\vec{i}-21\vec{j}\\ &\textrm{d}.\quad \color{red}21\vec{i}-22\vec{j}\\ &\textrm{e}.\quad 22\vec{i}-23\vec{j}\\\\ &\textrm{Jawab}\\ &\begin{aligned}\textrm{Ingat}&\textrm{lah akan tigaan Pythagoras}\\ &\begin{cases} (3,4,5) &\Rightarrow 3^{2}+4^{2}=5^{2} \\ (5,12,13) & \Rightarrow 5^{2}+12^{2}=13^{2} \\ (8,15,17) &\Rightarrow \cdots \\ (20,21,29)&\Rightarrow \cdots \\ \vdots & dll \end{cases}\\ \textrm{Sehin}&\textrm{gga}\\ \textrm{yang}&\: \: \textrm{paling mungkin adalah}\: : \: \\ &=\sqrt{20^{2}+21^{2}}=\sqrt{400+441}\\ &=\sqrt{841}\\ &=\color{red}29 \end{aligned} \end{array}$

Proyeksi Ortogonal Suatu Vektor

 

$\color{blue}\begin{aligned}\textrm{A}.\quad&\textrm{Panjang Proyeksi Ortogonal Suatu}\\ &\textrm{Vektor pada vektor lain} \end{aligned}$.

Perhatikanlah ilstrasi gambar yang dibentuk dari dua vektor berikut

Pada gambar di atas 

$\begin{array}{|c|c|}\hline \triangle \textrm{OAC}&\angle \left ( \overrightarrow{a},\overrightarrow{b} \right )\\\hline \begin{aligned}\cos \theta &=\displaystyle \frac{\left | \overrightarrow{c} \right |}{\left | \overrightarrow{a} \right |}\\ \Leftrightarrow \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \: \: ........(1) \end{aligned}&\begin{aligned}\cos \theta &=\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |}\: \: ........(2) \end{aligned}\\\hline \end{array}$.

$\begin{aligned}\textrm{Dari}\: \: (1)\: \: &\textrm{dan} \: \: (2)\: \: \textrm{diperoleh}\\ \left | \overrightarrow{c} \right |&=\left | \overrightarrow{a} \right |\cos \theta \\ &=\left | \overrightarrow{a} \right |\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{a} \right |\left | \overrightarrow{b} \right |} \right )\\ &=\color{red}\left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \end{aligned}$

$\color{blue}\begin{aligned}\textrm{B}.\quad&\textrm{Proyeksi Ortogonal Suatu Vektor}\\ &\textrm{pada vektor lain} \end{aligned}$.

$\begin{array}{|c|}\hline {\textrm{Perhatikan pula misal}\: \: \hat{c}\: \: \textrm{adalah vektor satuan dari}\: \: \overrightarrow{c}\: \: \textrm{dan}\: \: \overrightarrow{b},}\\\hline \begin{aligned}\textrm{maka}\: \: \: \overrightarrow{c}&=\left | \overrightarrow{c} \right |\hat{c} \end{aligned}\qquad\qquad \textrm{dan}\qquad\qquad \begin{aligned}\overrightarrow{b}&=\left | \overrightarrow{b} \right |\hat{b}=\left | \overrightarrow{b} \right |\hat{c} \end{aligned}\\\hline {\begin{aligned}\textrm{Sehingga}&\: \: \textbf{proyeksi ortogonal vektor}\: \: \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{adalah}:\\ \overrightarrow{c}&=\left | \overrightarrow{c} \right |\hat{b}\\ &=\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right )\left ( \displaystyle \frac{\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right )\\ &=\left (\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |^{2}} \right )\overrightarrow{b} \end{aligned}}\\\hline \end{array}$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$.

$\begin{array}{ll}\\ 1.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=\begin{pmatrix} 3\\ 2 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} -2\\ 1 \end{pmatrix}.\\ & \textrm{Tentukanlah proyeksi ortogonal vektor}\\ &\overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{dan tentukanlah panjangnya} \\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Misalkan}&\: \: \overrightarrow{c}\: \: \textrm{adalah vektor proyeksi yang dimaksud, }\\ &\textrm{maka}\\ \overrightarrow{c}&=\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |^{2}} \right )\overrightarrow{b}\\ &=\displaystyle \frac{\begin{pmatrix} 3\\ 2 \end{pmatrix}.\begin{pmatrix} -2\\ 1 \end{pmatrix}}{(-2)^{2}+1^{2}}.\overrightarrow{b}=\frac{3.(-2)+2.1}{4+1}\begin{pmatrix} -2\\ 1 \end{pmatrix}\\ &=-\frac{4}{5}\begin{pmatrix} -2\\ 1 \end{pmatrix}=\begin{pmatrix} \frac{8}{5}\\ -\frac{4}{5} \end{pmatrix}\quad \textbf{atau}\\ &=\color{red}\frac{8}{5}\bar{i}-\frac{4}{5}\bar{j} \end{aligned}\\ &\begin{aligned}\textrm{Dan panjang}\: \: &\textrm{vektor proyeksi yang dimaksud adalah}:\\ \left |\overrightarrow{c} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ 2 \end{pmatrix}.\begin{pmatrix} -2\\ 1 \end{pmatrix}}{\sqrt{(-2)^{2}+1^{2}}} \right |=\left |\frac{3.(-2)+2.1}{\sqrt{4+1}} \right |\\ &=\left |-\frac{4}{\sqrt{5}} \right |\\ &=\displaystyle \frac{4}{\sqrt{5}}=\color{red}\frac{4}{5}\sqrt{5} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=\begin{pmatrix} -5\\ 4 \end{pmatrix}\: \: \textrm{dan}\: \: \overrightarrow{b}=\begin{pmatrix} 2\\ 6 \end{pmatrix}.\\ & \textrm{Tentukanlah proyeksi ortogonal vektor}\\ &\overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\: \: \textrm{dan tentukanlah panjangnya} \\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{Misalkan}&\: \: \overrightarrow{c}\: \: \textrm{adalah vektor proyeksi yang dimaksud, }\\ &\textrm{maka}\\ \overrightarrow{c}&=\left ( \displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |^{2}} \right )\overrightarrow{b}\\ &=\displaystyle \frac{\begin{pmatrix} -5\\ 4 \end{pmatrix}.\begin{pmatrix} 2\\ 6 \end{pmatrix}}{2^{2}+6^{2}}.\overrightarrow{b}=\frac{(-5).2+4.6}{4+36}\begin{pmatrix} 2\\ 6 \end{pmatrix}\\ &=\frac{7}{20}\begin{pmatrix} 2\\ 6 \end{pmatrix}=\begin{pmatrix} \frac{7}{10}\\ \frac{21}{10} \end{pmatrix}\quad \textbf{atau}\\ &=\color{red}\frac{7}{10}\bar{i}+\frac{21}{10}\bar{j} \end{aligned}\\ &\begin{aligned}\textrm{Dan panjang}\: \: &\textrm{vektor proyeksi yang dimaksud adalah}:\\ \left |\overrightarrow{c} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} -5\\ 4 \end{pmatrix}.\begin{pmatrix} 2\\ 6 \end{pmatrix}}{\sqrt{2^{2}+6^{2}}} \right |=\left |\frac{(-5).2+4.6}{\sqrt{4+36}} \right |\\ &=\left |\frac{14}{\sqrt{40}} \right |\\ &=\displaystyle \frac{14}{2\sqrt{10}}=\color{red}\frac{7}{10}\sqrt{10} \end{aligned} \end{array}$.

$.\quad \color{red}\textrm{Coba bandingkan dengan vektor di dimensi tiga berikut}$

$\begin{array}{ll}\\ 3.&\textrm{Diketahui vektor}\: \: \overrightarrow{a}=3\bar{i}-2\bar{j}+2\bar{k}\: \: \textrm{dan}\: \: \overrightarrow{b}=2\bar{i}-2\bar{j}+\bar{k}\\ &\textrm{Tentukanlah panjang vektor proyeksi ortogonal}\\ &\textrm{a}.\quad \overrightarrow{a}\: \: \textrm{pada}\: \: \overrightarrow{b}\qquad\qquad\qquad \textrm{b}.\quad \overrightarrow{a}\: \: \textrm{pada}\: \: \left ( \overrightarrow{a}+\overrightarrow{b} \right ) \\\\ &\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad\textrm{Misal}&\textrm{kan}\: \: \overrightarrow{c}\: \: \textrm{adalah vektor proyeksi yang dimaksud,}\\ & \textrm{maka panjanynya}\\ \left |\overrightarrow{c} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\overrightarrow{b}}{\left | \overrightarrow{b} \right |} \right | \\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ -2\\ 2 \end{pmatrix}.\begin{pmatrix} 2\\ -2\\ 1 \end{pmatrix}}{\sqrt{2^{2}+(-2)^{2}+1^{2}}} \right |=\left |\frac{3.2+(-2).(-2)+2.1}{\sqrt{4+4+1}} \right |\\ &=\left | \displaystyle \frac{12}{3} \right |=\color{red}4 \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad\textrm{Misal}&\textrm{kan}\: \: \overrightarrow{f}\: \: \textrm{adalah vektor proyeksi yang dimaksud,}\\ & \textrm{maka panjanynya}\\ \left |\overrightarrow{f} \right |&= \left |\displaystyle \frac{\overrightarrow{a}\left (\overrightarrow{a}+\overrightarrow{b} \right )}{\left |\overrightarrow{a}+ \overrightarrow{b} \right |} \right |\\ &=\left |\displaystyle \frac{\begin{pmatrix} 3\\ -2\\ 2 \end{pmatrix}.\begin{pmatrix} 3+2\\ -2+(-2)\\ 2+1 \end{pmatrix}}{\sqrt{(3+2)^{2}+(-2+(-2))^{2}+(2+1)^{2}}} \right |\\ &=\left |\frac{3.5+(-2).(-4)+2.3}{\sqrt{25+16+9}} \right |\\ &=\left |\frac{29}{\sqrt{50}} \right |\\ &=\displaystyle \frac{29}{5\sqrt{2}}=\color{red}\frac{29}{10}\sqrt{5} \end{aligned} \end{array}$