Question

The distance between the line $r=2\hat{i}-2\hat{j}+3\hat{k}+\lambda (\hat{i}-\hat{j}+4\hat{k})$ and the plane $r.(\hat{i}+5\hat{j}+\hat{k})=5$ is

• A. $\frac{10}{9}$
• B. $\frac{10}{3\sqrt{3}}$
• C. $\frac{10}{3}$
• D. None of these

Answer 1

The correct option is B $\frac{10}{3\sqrt{3}}$

The given line is $r=a+tb$
where $a=2\hat{i}-2\hat{j}+3\hat{k},b=\hat{i}-\hat{j}+4\hat{k}$ and given plane is $r.n=p,$
where $n=\hat{i}+5\hat{j}+\hat{k},p=5.$
Since $b.n=1-5+4=0,$
$\therefore$ given line is parallel to the given plane.

$\therefore$ the distance between the line and the plane is equal to length of the perpendicular from the point $a=2\hat{i}-2\hat{j}+3\hat{k}$ on the given plane.
$\therefore$ required distance $=\left|\frac{(2i-2j+3k).(i+5j+k)-5}{\sqrt{1+25+1}}\right|=\left|\frac{2-10+3-5}{\sqrt{27}}\right|=\frac{10}{3\sqrt{3}}.$