Question

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

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

Answer 1

The correct option is A

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

Explanation for the correct option:

Find the distance between the line vector and plane vector:

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

Compared given equation with the standard form of the line $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ which is in vector form.

We get,

$\overrightarrow{a}=2\hat{i}-2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=(\hat{i}-\hat{j}+4\hat{k})$.

Similarly, compare the given plane vector to its standard form.

Given plane vector as $\overrightarrow{r·}\left(\hat{i}+5\hat{j}+\hat{k}\right)=5$

Standard form of the plane vector as $\overrightarrow{r}.\overrightarrow{n}=d$ .

Then,

$\overrightarrow{n}=\left(\hat{i}+5\hat{j}+\hat{k}\right)$ and $d=5$.

we know that.

The formula of the distance between the line and a plane vector is

$D=\left|\frac{\overrightarrow{a}.\overrightarrow{n}-d}{\left|\overrightarrow{n}\right|}\right|$

Substitute all the values,

We get,

$$\begin{gathered} D=\left|\frac{\left(2\hat{i}-2\hat{j}+3\hat{k}\right)·\left(\hat{i}+5\hat{j}+\hat{k}\right)-5}{\sqrt{1+25+1}}\right| \\ D=\left|\frac{-10}{3\sqrt{3}}\right| \\ D=\frac{10}{3\sqrt{3}} \end{gathered}$$

Hence, option (A) is the correct answer.