Question

The distance of the point $(1,-5,9)$, from the planer. $(\hat{i}-\hat{j}+\hat{k})=5$ measured long the line $r=\hat{i}+\hat{j}+\hat{k}$ is

• A. $3\sqrt{5}$
• B. $10\sqrt{3}$
• C. $5\sqrt{3}$
• D. $3\sqrt{10}$

Answer 1

Answer: (B)

$10\sqrt{3}$

Let $Q$ be the point where the line through $P(1,-5,9)$ and parallel to $r=\hat{i}+\hat{j}+\hat{k}$ meets the plane $r(\hat{i}-\hat{j}+\hat{k})=5$.
Equation of $PQ$, is $\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$.
Therefore, $x=\lambda +1$,
$y=\lambda -5$
and $z=\lambda +9$ lies on the plane $x-y+z=5$
$\Rightarrow \lambda +1-\lambda +5+\lambda +9=5$
$\Rightarrow \lambda =-10$
Therefore, co-ordinates of $Q$ are $Q(-9,-15,-1)$.
$\Rightarrow PQ=\sqrt{{10}^2+{10}^2+{10}^2}=10\sqrt{3}$.