Question

The distance from the point $-\hat{i}+2\hat{j}+6\hat{k}$ to the straight line passing through the point with position vector $2\hat{i}+3\hat{j}-4\hat{k}$ and parallel to the vectors $6\hat{i}+3\hat{j}-4\hat{k}$ is

• A. $10$
• B. $7$
• C. $5$
• D. $3$

Answer 1

Answer: (B)

$7$

Let $A=(2+6t,3+3t,-4-4t)$ is the point on the line which is least distance from the point $B=(2,3,-4)$.

Then the vector $BA=(3+6t)i+(2+3t)j+(-10-4t)k$ is perpendicular to $6i+3j-4k$.

$\Rightarrow (3+6t)\times 6+(2+3t)3-(10-4t)4=0$

$\Rightarrow t=-1$

$\Rightarrow A=(-4,0,0)$.

The distance from A to $(-1,2,6)$ is $\sqrt{(-3{)}^2+2^2+6^2}=7$.