Question

Find the distance of the point (2, 4, −1) from the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$. [NCERT EXEMPLAR]

Answer 1

We know that the distance d from point P to the line l having equation $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$ is given by $d=\frac{\left|\overrightarrow{b}\times \overrightarrow{\mathrm{PQ}}\right|}{\left|\overrightarrow{b}\right|}$, where Q is any point on the line l.

The equation of the given line is $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$.

Let P(2, 4, −1) be the given point. Now, in order to find the required distance we need to find a point Q on the given line.

The given line passes through the point (−5, −3, 6). So, take this point as the required point Q(−5, −3, 6).

Also, the line is parallel to the vector $\overrightarrow{b}=\hat{i}+4\hat{j}-9\hat{k}$.

Now, $\overrightarrow{\mathrm{PQ}}=\left(-5\hat{i}-3\hat{j}+6\hat{k}\right)-\left(2\hat{i}+4\hat{j}-\hat{k}\right)=-7\hat{i}-7\hat{j}+7\hat{k}$

$$\begin{gathered} \therefore \overrightarrow{b}\times \overrightarrow{\mathrm{PQ}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 4& -9\\ -7& -7& 7\end{array}\right|=-35\hat{i}+56\hat{j}+21\hat{k} \\ \Rightarrow \left|\overrightarrow{b}\times \overrightarrow{\mathrm{PQ}}\right|=\sqrt{{\left(-35\right)}^2+{56}^2+{21}^2}=\sqrt{1225+3136+441}=\sqrt{4802}=49\sqrt{2} \end{gathered}$$

Let d be the required distance.

$\therefore d=\frac{\left|\overrightarrow{b}\times \overrightarrow{\mathrm{PQ}}\right|}{\left|\overrightarrow{b}\right|}=\frac{49\sqrt{2}}{\sqrt{1+16+81}}=\frac{49\sqrt{2}}{\sqrt{98}}=\frac{49\sqrt{2}}{7\sqrt{2}}=7$

Thus, the distance of the given point from the given line is 7 units.