Question

The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is

• A. $3\sqrt{10}$
• B. $10\sqrt{3}$
• C. $\frac{10}{\sqrt{3}}$
• D. $\frac{20}{3}$

Answer 1

The correct option is B

$10\sqrt{3}$

Equation of line passing through the point (1, -5, 9) and parallel to x = y = z is
$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=\lambda$
Thus, any point on this line is of the form
$\lambda +1,\lambda -5,\lambda +9$.
Now, if $P(\lambda +1,\lambda -5,\lambda +9)$ is the point of intersection of line and plane, then
$(\lambda +1)-(\lambda -5)+\lambda +9=5$
$\Rightarrow \lambda +15=5\Rightarrow \lambda =-10$
$\therefore$ Coordintes of point P are (-9, -15, -1).
Hence, required distance
$=\sqrt{(1+9{)}^2+(-5+15{)}^2+(9+1{)}^2}$
$=\sqrt{{10}^2+{10}^2+{10}^2}=10\sqrt{3}$