Question

The distance of the point $(1,0,2)$ from the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ and the plane

• A. $2\sqrt{1}4$
• B. $8$
• C. $3\sqrt{2}1$
• D. $13$

Answer 1

Answer: (D)

Given equation of line is
$\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda$ [say] ...(i)
and equation of plane is
$x-y+z=16$
Any point on the line (i) is
$(3\lambda +2,4\lambda -1,12\lambda +2)$
Let this point of intersection of the line and plane.
$\therefore (3\lambda +2)-(4\lambda -1)+(12\lambda +2)=16$
$11\lambda +5=16$
$11\lambda =11\Rightarrow \lambda =1$
So, the point of intersection is $(5,3,14)$.
Now, distance between the points $(1,0,2)$ and $(5,3,14)$
$=\sqrt{(5-1{)}^2+(3-0{)}^2+(14-2{)}^2}$
$=\sqrt{16+9+144}=\sqrt{169}=13$