Question

The plane $2x-3y+z+6=0$ divides the line segment joining $(2,4,16)$ and $(3,5,-4)$ in the ratio

• A. $4:5$
• B. $4:7$
• C. $2:1$
• D. $1:2$

Answer 1

The correct option is C $2:1$

Let the line segment joining points $P(2,4,16)$ and $Q(3,5,-4)$ be divided by the given plane in the ratio $k:1$ at the point $R.$

Then co-ordinates of $R$ are $(\frac{3k+2}{k+1},\frac{5k+4}{k+1},\frac{-4k+16}{k+1})$

Since $R$ lies on the plane $2x-3y+z+6=0,$

We have, $2\left(\frac{3k+2}{k+1}\right)-3\left(\frac{5k+4}{k+1}\right)+\left(\frac{-4k+16}{k+1}\right)+6=0$

$\Rightarrow 2(3k+2)-3(5k+4)+(-4k+16)+6(k+1)=0$

$\Rightarrow -7k+14=0\Rightarrow k=2$

$\therefore PQ$ is divided by the given plane in the ratio $k:1$ i.e., $2:1$