Question

Find the perpendicular distance of a point $\left(1,0,0\right)$ from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$.

Answer 1

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 $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ are given by

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

$\Rightarrow x-1=2\lambda ,y+1=-3\lambda ,z+10=8\lambda$

$\Rightarrow x=2\lambda +1,y=-3\lambda -1,z=8\lambda -10$

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

Direction ratios of $PL$ are proportional to $2\lambda +1-1,-3\lambda -1-0,8\lambda -10-0$

or $2\lambda ,-3\lambda -1,8\lambda -10$

Direction ratios of the given line are proportional to $2,-3,8$

Since $PL$ is perpendicular to the given line.

$\therefore 2\left(2\lambda \right)-3(-3\lambda -1)+8(8\lambda -10)=0$

$\Rightarrow 4\lambda +9\lambda +3+64\lambda -80=0$

$\Rightarrow 73\lambda -73=0$

$\Rightarrow 73\lambda =73$

$\Rightarrow \lambda =\frac{73}{73}=1$

Put $\lambda =1$ in $\left(1\right)$, we obtain the coordinates of $L$

$(2+1,-3-1,8-10)$ or $(3,-4,-2)$

$\therefore PL=\sqrt{{(1-3)}^2+{(0+4)}^2+{(0+2)}^2}=\sqrt{4+16+4}=\sqrt{24}=\sqrt{4\times 6}=2\sqrt{6}$units.

Hence, the length of the perpendicular is $2\sqrt{6}$ units.