Question

If the vectors $\overrightarrow{a}=2\hat{i}+3\hat{j}-6\hat{k}and\hat{b}=x\hat{i}-\hat{j}+2\hat{k}$ are parallel, then x =

• A. $2$
• B. $\frac{2}{3}$
• C. $-\frac{2}{3}$
• D. $\frac{1}{3}$

Answer 1

Answer: (C)

$-\frac{2}{3}$

Two vectors $\overrightarrow{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\overrightarrow{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$ are parallel if

$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}=k$

where $k$ is a non-zero scalar.

Now, the given vectors $\overrightarrow{a}=2\hat{i}+3\hat{j}-6\hat{k}$ and $\overrightarrow{b}=x\hat{i}-1\hat{j}+2\hat{k}$ are parallel if

$\frac{2}{x}=\frac{3}{-1}=\frac{-6}{2}=k$

it implies,

$\frac{2}{x}=-3$

$x=\frac{-2}{3}$