Question

Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane $x-y+z=5$. [CBSE 2014, 2015]

Answer 1

The equation of the line passing through the points A(2, −1, 2) and B(5, 3, 4) is given by

$$\begin{gathered} \frac{x-2}{5-2}=\frac{y-\left(-1\right)}{3-\left(-1\right)}=\frac{z-2}{4-2} \\ \mathrm{Or}\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2} \end{gathered}$$

The coordinates of any point on the line

$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda \left(\mathrm{say}\right)$ are $\left(3\lambda +2,4\lambda -1,2\lambda +2\right)$ .....(1)

If it lies on the plane $x-y+z=5$, then

$$\begin{gathered} 3\lambda +2-\left(4\lambda -1\right)+2\lambda +2=5 \\ \Rightarrow \lambda +5=5 \\ \Rightarrow \lambda =0 \end{gathered}$$

Putting $\lambda =0$ in (1), we get (2, −1, 2) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (−1, −5, −10) and (2, −1, 2)

$$\begin{gathered} =\sqrt{{\left(2+1\right)}^2+{\left(-1+5\right)}^2+{\left(2+10\right)}^2} \\ =\sqrt{9+16+144} \\ =\sqrt{169} \\ =13\mathrm{units} \end{gathered}$$