Question

The distance of the line $\overrightarrow{r}=2\hat{i}-2\hat{j}+3\hat{k}+\lambda \left(\hat{i}-\hat{j}+4\hat{k}\right)$ from the plane $\overrightarrow{r}·\left(\hat{i}+5\hat{j}+\hat{k}\right)=5$ is
(a) $\frac{5}{3\sqrt{3}}$

(b) $\frac{10}{3\sqrt{3}}$

(c) $\frac{25}{3\sqrt{3}}$

(d) None of these

Answer 1

(b) $\frac{10}{3\sqrt{3}}$

$$\begin{gathered} \text{The given line passes through the point whose position vector is}\overrightarrow{a}\text{=2}\hat{i}\text{-2}\hat{j}\text{+3}\hat{k}\text{.} \\ \text{We know that the perpendicular distance of a point}\mathit{\text{P}}\text{of position vector}\overrightarrow{a}\text{from the plane}\overrightarrow{r}\text{.}\overrightarrow{n}\text{=}\mathit{\text{d}}\text{is given by} \\ p=\frac{\left|\overrightarrow{a}.\overrightarrow{n}-d\right|}{\left|\overrightarrow{n}\right|} \\ \text{Here,}\overrightarrow{a}=2\hat{i}\text{-2}\hat{j}\text{+3}\hat{k};\overrightarrow{n}=\hat{i}+5\hat{j}+\hat{k};d=5 \\ \text{So, the required distance}\mathit{\text{p}}\text{is given by} \\ p=\frac{\left|\left(2\hat{i}\text{-2}\hat{j}\text{+3}\hat{k}\right).\left(\hat{i}+5\hat{j}+\hat{k}\right)-5\right|}{\left|\hat{i}+5\hat{j}+\hat{k}\right|} \\ =\frac{\left|2-10+3-5\right|}{\sqrt{1+25+1}} \\ =\frac{\left|-10\right|}{\sqrt{27}} \\ =\frac{10}{3\sqrt{3}}\text{units} \end{gathered}$$