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  $$x-y+z=5$$ . 

 

Answer 1

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

$$\frac{x-2}{5-2}=\frac{y-(-1)}{3-(-1)}=\frac{z-2}{4-2}$$
$$\text{ Or }\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}$$

The coordinates of any point on the line 

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

$$(3\lambda +2,4\lambda -1,2\lambda +2)$$

 .........(1)

If it lies on the plane

$$x-y+z=5$$ , then 

$$3\lambda +2-(4\lambda -1)+2\lambda +2=5$$
$$\Rightarrow \lambda +5=5$$
$$\Rightarrow \lambda =0$$

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)

$$=\sqrt{{(2+1)}^2+{(-1+5)}^2+{(2+10)}^2}$$
$$=\sqrt{9+16+144}$$
$$=\sqrt{169}$$
$$=13\text{ units}$$