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]

Open in App
Solution

## The equation of the line passing through the points A(3, −4, −5) and B(2, −3, 1) is given by $\frac{x-3}{2-3}=\frac{y-\left(-4\right)}{-3-\left(-4\right)}=\frac{z-\left(-5\right)}{1-\left(-5\right)}\phantom{\rule{0ex}{0ex}}\mathrm{Or}\frac{x-3}{-1}=\frac{y+4}{1}=\frac{z+5}{6}$ 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 $2\left(-\lambda +3\right)+\left(\lambda -4\right)+\left(6\lambda -5\right)=7\phantom{\rule{0ex}{0ex}}⇒5\lambda -3=7\phantom{\rule{0ex}{0ex}}⇒5\lambda =10\phantom{\rule{0ex}{0ex}}⇒\lambda =2$ 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) $=\sqrt{{\left(3-1\right)}^{2}+{\left(4+2\right)}^{2}+{\left(4-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{4+36+9}\phantom{\rule{0ex}{0ex}}=\sqrt{49}\phantom{\rule{0ex}{0ex}}=7\mathrm{units}$

Suggest Corrections
1
Related Videos
Line in Three Dimensional Space
MATHEMATICS
Watch in App