Question

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

• A. $\frac{10}{3\sqrt{3}}$
• B. $\frac{10}{9}$
• C. $\frac{10}{3}$
• D. $\frac{3}{10}$

Answer 1

Answer: (A)

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

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) ...(1)

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 (1).

To know intersection of (1) 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 $=\sqrt{{\left(\frac{10}{27}\right)}^2+{\left(\frac{50}{27}\right)}^2+{\left(\frac{10}{27}\right)}^2}=\sqrt{\frac{2700}{{27}^2}}=\frac{10}{3\sqrt{3}}$

Hence, option 'A' is correct.