Question

The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is

• A. $1$
• B. $19$
• C. $155$
• D. $\sqrt{271}$

Answer 1

Answer: (A)

$1$

Equation of the line through $(1,-2,3)$ parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is
$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r$ (say) ...(1)
Then any point on (1) is $(2r+1,3r-2,-6r+3)$
If this point lies on the plane $x-y+z=5$ then $(2r+1)-(3r-2)+(-6r+3)=5$
$\Rightarrow -7r+6=5\Rightarrow r=\frac{1}{7}$
Hence, the point is $(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$
Distance between $(1,-2,3)$ and $(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$
$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{\frac{49}{49}}=1$