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)

$\frac {\sqrt {293} } {7}$

Answer 2

Let $\left(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}=t\right)$
$\Rightarrow(x, y, z)=(2 t+3,3 t+3,6 t-5)$
$\therefore$ d.r.?s of the line perpendicular to
$\left(\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}\right)$ and
joining $(2 t+3,3 t+3,6 t-5)$
and $(1,2,-4)$ is $(2 t+2,3 t+1,6 t-1)$
$\therefore 2(2 t+2)+3(3 t+1)+6(6 t-1)=0$
$\Rightarrow t =-1 / 49$
$\therefore$ Distance $=\sqrt{(2t+2)^2+(3t+1)^2+(6t-1)^2}$
$= \sqrt{49t^2+2t+6}$
$= \sqrt{\frac{1}{49}-\frac{2}{49}+6}=\frac{\sqrt{293}}{7}$