Question

The distance of the point of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and the plane $x+y+z=17$ from the point $(3,4,5)$ is given by

• A. $3$
• B. $\frac{3}{2}$
• C. $\sqrt{3}$
• D. None of these

Answer 1

The correct option is A $3$

Equation of the line is given as

$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda$ (say) ……………$\left(1\right)$

$\therefore x=\lambda +3$

$y=2\lambda +4$

$z=2\lambda +5$

$\Rightarrow (\lambda +3,2\lambda +4,2\lambda +5)$ is any point on the line.

Since the line $\left(1\right)$ interest with the plane $x+y+z=17$

$\therefore (\lambda +3)+(2\lambda +4)+(2\lambda +5)=17$

$\Rightarrow 5\lambda +12=17$

$\Rightarrow 5\lambda =5$

$\Rightarrow \lambda =1$

$\therefore$ Point of intersection $=(1+3,2+4,2+5)$

$=(4,6,7)$

$\therefore$ Distance between $(4,6,7)$ and $(3,4,5)$

$=\sqrt{(4-3{)}^2+(6-4{)}^2+(7-5{)}^2}$

$=\sqrt{1^2+2^2+2^2}$

$=\sqrt{1+4+4}$

$=\sqrt{9}$

$=3$.