Question

Find the perpendicular distance of the point (1, 0, 0) from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}.$ Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

Answer 1

Let the point (1, 0, 0) be P and the point through which the line passes be Q (1, $-$1, $-$10).
The line is parallel to the vector $\overrightarrow{b}=2\hat{i}-3\hat{j}+8\hat{k}$.

Now,

$\overrightarrow{PQ}=0\hat{i}-\hat{j}-10\hat{k}$
$$\begin{gathered} \therefore \overrightarrow{b}\times \overrightarrow{PQ}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -3& 8\\ 0& -1& -10\end{array}\right| \\ =38\hat{i}+20\hat{j}-2\hat{k} \\ \Rightarrow \left|\overrightarrow{b}\times \overrightarrow{PQ}\right|=\sqrt{{38}^2+{20}^2+2^2} \\ =\sqrt{1444+400+4} \\ =\sqrt{1848} \\ d=\frac{\left|\overrightarrow{b}\times \overrightarrow{PQ}\right|}{\left|\overrightarrow{b}\right|} \\ =\frac{\sqrt{1848}}{\sqrt{77}} \\ =\sqrt{24} \\ =2\sqrt{6} \end{gathered}$$


Let L be the foot of the perpendicular drawn from the point P (1, 0, 0) to the given line.



The coordinates of a general point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ are given by

$$\begin{gathered} \frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}=\lambda \\ \Rightarrow x=2\lambda +1 \\ y=-3\lambda -1 \\ z=8\lambda -10 \end{gathered}$$

Let the coordinates of L be $\left(2\lambda +1,-3\lambda -1,8\lambda -10\right)$.

The direction ratios of PL are proportional to $2\lambda +1-1,-3\lambda -1-0,8\lambda -10-0,\mathrm{i}.\mathrm{e}.2\lambda ,-3\lambda -1,8\lambda -10$.

The direction ratios of the given line are proportional to 2 ,$-$3, 8, but PL is perpendicular to the given line.

$$\begin{gathered} \therefore 2\left(2\lambda \right)-3\left(-3\lambda -1\right)+8\left(8\lambda -10\right)=0 \\ \Rightarrow \lambda =1 \end{gathered}$$

Substituting $\lambda =1$ in $\left(2\lambda +1,-3\lambda -1,8\lambda -10\right)$, we get the coordinates of L as (3, $-$4, $-$2).

Equation of the line PL is given by

$$\begin{gathered} \frac{x-1}{3-1}=\frac{y-0}{-4-0}=\frac{z-0}{-2-0} \\ =\frac{x-1}{1}=\frac{y}{-2}=\frac{z}{-1} \end{gathered}$$

$\Rightarrow \overrightarrow{r}=\hat{i}+\lambda \left(\hat{i}-2\hat{j}-\hat{k}\right)$