Question

Given $\overrightarrow{\alpha }=3\hat{i}+\hat{j}+2\hat{k},\overrightarrow{\beta }=\hat{i}-2\hat{j}-4\hat{k}$ are the position vectors of the points $A$ and $B$. Then the distance of the point $-\hat{i}+\hat{j}+\hat{k}$ from the passing through $B$ and perpendicular to $AB$ is

• A. $-7$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is A $-7$

We have,

$\overrightarrow{\alpha }=3\hat{i}+\hat{j}+2\hat{k}$

$\overrightarrow{\beta }=\hat{i}-2\hat{j}-4\hat{k}$

So,

$\overrightarrow{AB}=\overrightarrow{\beta }-\overrightarrow{\alpha }$

$\overrightarrow{AB}=(\hat{i}-2\hat{j}-4\hat{k})-(3\hat{i}+\hat{j}+2\hat{k})$

$\overrightarrow{AB}=\hat{i}-2\hat{j}-4\hat{k}-3\hat{i}-\hat{j}-2\hat{k}$

$\overrightarrow{AB}=-2\hat{i}-3\hat{j}-6\hat{k}$

Now, equation of plane passing through the point and perpendiular vector is

So,

$(-2\hat{i}-3\hat{j}-6\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=(-2\hat{i}-3\hat{j}-6\hat{k}).(-\hat{i}+\hat{j}+\hat{k})$

$\Rightarrow 2x+3y+6z=2-3-6$

$\Rightarrow 2x+3y+6z=-7$

Hence, this is the answer.