Question

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

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

Answer 1

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

Given that, $\overline{r}=2\hat{i}-2\hat{j}+3\hat{k}+\lambda (\hat{i}-\hat{j}+4\hat{k})$

plane is $\overline{r}.(\hat{i}+5\hat{j}+\hat{k})=5$

As the line is parallel to plane $[\because (\hat{i}-\hat{j}+4\hat{k}).(\hat{i}+5\hat{j}+\hat{k})=0]$

Any point on the line is of form $(\lambda +2,-\lambda -2,4\lambda +3)$

For $\lambda =0$ , point on this line is (2, -2, 3) and its distance from $r.(\hat{i}+5\hat{j}+\hat{k})=5$ is

$d=\left|\frac{2+5(-2)+3\left(1\right)-5}{\sqrt{1^2+5^2+1^2}}\right|$

$d=\left|\frac{2-10+3-5}{\sqrt{27}}\right|$

$d=\left|\frac{-10}{3\sqrt{3}}\right|$

As distance can"t be negative, we take modulas,

$\therefore$ Distance between line and plane is $\frac{10}{3\sqrt{3}}$