Question

Find the points on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance of 5 units from the point P (1, 3, 3).

Answer 1

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

$$\begin{gathered} \frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda \\ \Rightarrow x=3\lambda -2,y=2\lambda -1,z=2\lambda +3...\left(1\right) \end{gathered}$$

Let the coordinates of the desired point be $\left(3\lambda -2,2\lambda -1,2\lambda +3\right)$.

The distance between this point and (1, 3, 3) is 5 units.

$$\begin{gathered} \therefore \sqrt{{\left(3\lambda -2-1\right)}^2+{\left(2\lambda -1-3\right)}^2+{\left(2\lambda +3-3\right)}^2}=5 \\ \Rightarrow {\left(3\lambda -3\right)}^2+{\left(2\lambda -4\right)}^2+{\left(2\lambda \right)}^2=25 \\ \Rightarrow 17{\lambda }^2-34\lambda =0 \\ \Rightarrow \lambda \left(\lambda -2\right)=0 \\ \Rightarrow \lambda =0\mathrm{or}2 \end{gathered}$$

Substituting the values of $\lambda$ in (1), we get the coordinates of the desired point as ($-$2, $-$1, 3) and (4, 3, 7).