Question

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, −4, −5) and B(2, −3, 1) intersects the plane 2x + y + z = 7. [CBSE 2015]

Answer 1

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

$$\begin{gathered} \frac{x-3}{2-3}=\frac{y-\left(-4\right)}{-3-\left(-4\right)}=\frac{z-\left(-5\right)}{1-\left(-5\right)} \\ \mathrm{Or}\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6} \end{gathered}$$

The coordinates of any point on the line

$\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \left(\mathrm{say}\right)$ are $\left(-\lambda +3,\lambda -4,6\lambda -5\right)$ .....(1)

If it lies on the plane 2x + y + z = 7, then

$$\begin{gathered} 2\left(-\lambda +3\right)+\left(\lambda -4\right)+\left(6\lambda -5\right)=7 \\ \Rightarrow 5\lambda -3=7 \\ \Rightarrow 5\lambda =10 \\ \Rightarrow \lambda =2 \end{gathered}$$

Putting $\lambda =2$ in (1), we get (1, −2, 7) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (3, 4, 4) and (1, −2, 7)

$$\begin{gathered} =\sqrt{{\left(3-1\right)}^2+{\left(4+2\right)}^2+{\left(4-7\right)}^2} \\ =\sqrt{4+36+9} \\ =\sqrt{49} \\ =7\mathrm{units} \end{gathered}$$