Exercise Zone : Vektor

Berikut ini adalah kumpulan soal mengenai Vektor. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Diketahui A=(a,-2,3), B=(2,-1,1), dan C=(1,5,c). Agar vektor \overline{AB} tegak lurus pada \overline{BC}, maka nilai a-2c sama dengan ....

Ganesha Operation

$$\begin{array}{rl}\overline{AB}& =\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)-\left(\begin{array}{c}a\\ -2\\ 3\end{array}\right)\\ & =\left(\begin{array}{c}2-a\\ 1\\ -2\end{array}\right)\end{array}$$

$$\begin{array}{rl}\overline{BC}& =\left(\begin{array}{c}1\\ 5\\ c\end{array}\right)-\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}-1\\ 6\\ c-1\end{array}\right)\end{array}$$

\overline{AB} tegak lurus pada \overline{BC} maka
$$\begin{array}{rl}\overline{AB}\cdot \overline{BC}& =0\\ \left(\begin{array}{c}2-a\\ 1\\ -2\end{array}\right)\cdot \left(\begin{array}{c}-1\\ 6\\ c-1\end{array}\right)& =0\\ (2-a)(-1)+(1)(6)+(-2)(c-1)& =0\\ -2+a+6-2c+2& =0\\ a-2c& =\boxed{{\displaystyle \boxed{{\displaystyle -6}}}}\end{array}$$

Jadi, nilai a-2c sama dengan -6.

No.

P_1=(5,-2,1) dan P_2=(2,4,2) maka vektor 5\overrightarrow{P_1P_2} adalah....

2. 
   
3. 
   

5. 
   

Grup Facebook

$$\begin{array}{rl}5\overrightarrow{P_1{P}_2}& =5(\left(\begin{array}{c}2\\ 4\\ 2\end{array}\right)-\left(\begin{array}{c}5\\ -2\\ 1\end{array}\right))\\ & =5\left(\begin{array}{c}-3\\ 6\\ 1\end{array}\right)\\ & =\left(\begin{array}{c}-15\\ 30\\ 5\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle (-15,30,5)}}}}\end{array}$$

Jadi, 5\overrightarrow{P_1P_2}=(-15,30,5). TIDAK ADA PILIHAN JAWABAN

No.

Diketahui vektor {\vec{a}=(4,6)}, {\vec{b}=(3,4)} dan {\vec{c}=(p,0)}. Jika {\left|\vec{c}-\vec{a}\right|=10} maka cosinus sudut antara vector \vec{b} dan \vec{c} yang mungkin adalah

1. \dfrac25
2. \dfrac12
3. \dfrac35

4. \dfrac23
5. \dfrac34

CARA BIASA

$\begin{array}{rl}|\overrightarrow{c}-\overrightarrow{a}|& =10\\ \sqrt{(p-4{)}^2+(0-6{)}^2}& =10\\ \sqrt{p^2-8p+16+36}& =10\\ \sqrt{p^2-8p+52}& =10\\ p^2-8p+52& =100\\ p^2-8p-48& =0\\ (p+4)(p-12)& =0\end{array}$
p=-4 atau p=12

Untuk p=-4

\vec{c}=(-4,0)
$\begin{array}{rl}\cos (\overrightarrow{b},\overrightarrow{c})& ={\displaystyle \frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}}\\ & ={\displaystyle \frac{(3)(-4)+(4)(0)}{\sqrt{3^2+4^2}\sqrt{(-4{)}^2+0^2}}}\\ & ={\displaystyle \frac{-12+0}{\sqrt{9+16}\sqrt{16+0}}}\\ & ={\displaystyle \frac{-12}{\sqrt{25}\sqrt{16}}}\\ & ={\displaystyle \frac{-12}{(5)(4)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{3}{5}}}}}}\end{array}$

Untuk p=12

\vec{c}=(12,0)
$\begin{array}{rl}\cos (\overrightarrow{b},\overrightarrow{c})& ={\displaystyle \frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}}\\ & ={\displaystyle \frac{(3)(12)+(4)(0)}{\sqrt{3^2+4^2}\sqrt{{12}^2+0^2}}}\\ & ={\displaystyle \frac{36+0}{\sqrt{9+16}\sqrt{144+0}}}\\ & ={\displaystyle \frac{36}{\sqrt{25}\sqrt{144}}}\\ & ={\displaystyle \frac{36}{(5)(12)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3}{5}}}}}}\end{array}$

CARA CEPAT

$\begin{array}{rl}\cos (\overrightarrow{b},\overrightarrow{c})& ={\displaystyle \frac{\overrightarrow{b}\cdot \overrightarrow{c}}{\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|}}\\ & ={\displaystyle \frac{(3)(p)+(4)(0)}{\sqrt{3^2+4^2}\sqrt{p^2+0^2}}}\\ & ={\displaystyle \frac{3p+0}{\sqrt{9+16}\sqrt{p^2+0}}}\\ & ={\displaystyle \frac{3p}{\sqrt{25}\sqrt{p^2}}}\\ & ={\displaystyle \frac{3p}{\sqrt{25}|p|}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \pm {\displaystyle \frac{3}{5}}}}}}\end{array}$

Jadi, cosinus sudut antara vector \vec{b} dan \vec{c} yang mungkin adalah \dfrac35. JAWAB: C

No.

Diketahui \left|\vec{a}\right|=4, \left|\vec{b}\right|=5 serta \left|\vec{a}+\vec{b}\right|=6, tentukan nilai dari \left|\vec{a}-\vec{b}\right|

Grup Facebook

$$\begin{array}{rl}{|\overrightarrow{a}+\overrightarrow{b}|}^2+{|\overrightarrow{a}-\overrightarrow{b}|}^2& ={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2\\ 6^2+{|\overrightarrow{a}-\overrightarrow{b}|}^2& =4^2+5^2\\ 36+{|\overrightarrow{a}-\overrightarrow{b}|}^2& =16+25\\ {|\overrightarrow{a}-\overrightarrow{b}|}^2& =5\\ |\overrightarrow{a}-\overrightarrow{b}|& =\boxed{{\displaystyle \boxed{{\displaystyle \sqrt{5}}}}}\end{array}$$

Jadi, \left|\vec{a}-\vec{b}\right|=\sqrt5.

No.

Diketahui $\overrightarrow{PQ}=\left(\begin{array}{c}1\\ 0\end{array}\right)$ dan $\overrightarrow{PR}=\left(\begin{array}{c}2\\ 2\end{array}\right)$. Jika \overrightarrow{PS}=\dfrac14\overrightarrow{PQ}, maka \overrightarrow{RS}=

Grup Telegram

$\begin{array}{rl}\overrightarrow{PR}& =\left(\begin{array}{c}2\\ 2\end{array}\right)\\ \overrightarrow{r}-\overrightarrow{p}& =\left(\begin{array}{c}2\\ 2\end{array}\right)\\ \overrightarrow{p}-\overrightarrow{r}& =\left(\begin{array}{c}-2\\ -2\end{array}\right)\end{array}$

$\begin{array}{rl}\overrightarrow{PS}& ={\displaystyle \frac{1}{4}}\overrightarrow{PQ}\\ \overrightarrow{s}-\overrightarrow{p}& ={\displaystyle \frac{1}{4}}\left(\begin{array}{c}1\\ 0\end{array}\right)\\ & =\left(\begin{array}{c}{\displaystyle \frac{1}{4}}\\ 0\end{array}\right)\end{array}$

$\begin{array}{rl}\overrightarrow{RS}& =\overrightarrow{s}-\overrightarrow{r}\\ & =\overrightarrow{s}-\overrightarrow{p}+\overrightarrow{p}-\overrightarrow{r}\\ & =\left(\begin{array}{c}{\displaystyle \frac{1}{4}}\\ 0\end{array}\right)+\left(\begin{array}{c}-2\\ -2\end{array}\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \left(\begin{array}{c}-{\displaystyle \frac{7}{4}}\\ -2\end{array}\right)}}}}\end{array}$

Jadi, $\overrightarrow{RS}=\left(\begin{array}{c}-{\displaystyle \frac{7}{4}}\\ -2\end{array}\right)$.

No.

Diberikan {\vec{a}=2\vec{i}+3\vec{j}-2\vec{k}} dan {\vec{b}=3\vec{i}-3\vec{j}-4\vec{k}}. Hasil \vec{a}\cdot\vec{b} adalah

1. 3 satuan
2. 4 satuan
3. 5 satuan

4. 8 satuan
5. 10 satuan

Telegram

$\begin{array}{rl}\overrightarrow{a}\cdot \overrightarrow{b}& =(2)(3)+(3)(-3)+(-2)(-4)\\ & =6-9+8\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5}}}}\end{array}$

Jadi, \vec{a}\cdot\vec{b}=5 satuan. JAWAB: C

No.

Jika vektor {a=10i+6j-3k} dan {b=8i+3j+3k} serta {c=a-b}, maka vektor satuan yang searah dengan c adalah ....

1. \dfrac67i+\dfrac27j+\dfrac37k
2. \dfrac27i+\dfrac37j-\dfrac67k
3. \dfrac67i-\dfrac37j+\dfrac67k

4. \dfrac67i-\dfrac37j-\dfrac27k
5. -\dfrac27i+\dfrac67j-\dfrac37k

Grup Telegram

$\begin{array}{rl}c& =a-b\\ & =(10i+6j-3k)-(8i+3j+3k)\\ & =10i+6j-3k-8i-3j-3k\\ & =2i+3j-6k\end{array}$

$\begin{array}{rl}|c|& =\sqrt{2^2+3^2+(-6{)}^2}\\ & =\sqrt{4+9+36}\\ & =\sqrt{49}\\ & =7\end{array}$

$\begin{array}{rl}{e}_c& ={\displaystyle \frac{c}{|c|}}\\ & ={\displaystyle \frac{2i+3j-6k}{7}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2}{7}}i+{\displaystyle \frac{3}{7}}j-{\displaystyle \frac{6}{7}}k}}}}\end{array}$

Jadi, vektor satuan yang searah dengan c adalah \dfrac27i+\dfrac37j-\dfrac67k. JAWAB: B

No.

Tentukan panjang vektor p=(3,5,-4)

Brainly.co.id

$$\begin{array}{rl}|p|& =\sqrt{3^2+5^2+(-4{)}^2}\\ & =\sqrt{9+25+16}\\ & =\sqrt{50}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 5\sqrt{2}}}}}\end{array}$$

Jadi, panjang vektor p=(3,5,-4) adalah 5\sqrt2