Question

The distance of the point (1, 2, -4) from the line $\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}$ is

• A. $\frac{293}{7}$
• B. $\frac{\sqrt{293}}{7}$
• C. $\frac{293}{49}$
• D. $\frac{\sqrt{293}}{49}$

Answer 1

Answer: (B)

Let $\left(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}=t\right)$
$\Rightarrow (x,y,z)=(2t+3,3t+3,6t-5)$
$\therefore$ d.r.’s of the line perpendicular to $\left(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}\right)$ and
joining $(2t+3,3t+3,6t-5)$
and $(1,2,-4)$ is $(2t+2,3t+1,6t-1)$
$\therefore 2(2t+2)+3(3t+1)+6(6t-1)=0$
$\Rightarrow t=-1/49$
$\therefore$ Distance $=\sqrt{(2t+2{)}^2+(3t+1{)}^2+(6t-1{)}^2}$
$=\sqrt{49t^2+2t+6}$
$=\sqrt{\frac{1}{49}-\frac{2}{49}+6}=\frac{\sqrt{293}}{7}$