Question

If the line, $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane, $x+2y+3z=15$ at a point P, then the distance of P from the origin is

Answer 1

Answer: 4.5

Let point on line be $P(2k+1,3k-1,4k+2)$
Since, point $P$ lies on the plane $x+2y+3z=15$
$\therefore 2k+1+6k-2+12k+6=15$
$\Rightarrow k=\frac{1}{2}$
$\therefore P\equiv \left(2,\frac{1}{2},4\right)$
Then the distance of the point P from the origin is
$OP=\sqrt{4+\frac{1}{4}+16}=\frac{9}{2}$