Question

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

Answer 1

Solution.

The coordinates of the point $R$ which divides the line segment joining two points $P(x_1,y_1,z_1)\mathrm{and}Q(x_2,y_2,z_2)$ internally in the ratio $m:n$ are given by,

$\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n},\frac{m{z}_2+n{z}_1}{m+n}\right)$

The coordinates of the point $R$ which divides the line segment joining two points $(2,4,16)\mathrm{and}(3,5,-4)$ internally in the ratio $k:1$ are computed as,

$\left(\frac{3k+2}{k+1},\frac{5k+4}{k+1},\frac{-4k+16}{k+1}\right)$

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

$$\begin{gathered} \therefore 2\left(\frac{3k+2}{k+1}\right)-3\left(\frac{5k+4}{k+1}\right)+\left(\frac{-4k+16}{k+1}\right)+6=0 \\ \Rightarrow 6k+4-15k-12-4k+16+6k+6=0 \\ \Rightarrow -7k=-14 \\ \Rightarrow k=\frac{14}{7} \\ \Rightarrow k=2 \end{gathered}$$

Hence, the plane divides the given line segment in the ratio of $2:1$.