Question

The distance between the line $\frac{x-2}{1}=\frac{y+2}{-1}=\frac{z-3}{4}$and the plane $\overline{r}.(\overline{i}+\overline{5j}+\overline{1k})-5=0$ is

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

Answer 1

Answer: (C)

$\frac{10\sqrt{3}}{9}$

Clearly given line and plane are parallel.
Equation of line perpendicular to both is given by,
$\frac{x-2}{1}=\frac{y+2}{5}=\frac{z-3}{1}=k$(say) ...$\left(1\right)$
So any point on this line is given by, $P(k+2,5k-2,k+3)$
Distance between given line and plane will be along line $\left(1\right)$
To know intersection of $\left(1\right)$ and plane we should have,
$k+2+5(5k-2)+k+3=5\Rightarrow k=\frac{10}{27}$
Thus $P=(\frac{10}{27}-2,\frac{50}{27}-2,\frac{10}{27}+3)$
Hence, shortest distance is $=\sqrt{{\left(\frac{10}{27}\right)}^2+{\left(\frac{50}{27}\right)}^2+{\left(\frac{10}{27}\right)}^2}=\sqrt{\frac{2700}{{27}^2}}=\frac{10\sqrt{3}}{9}$