Question

Show that the plane ax + by + cz + d = 0 divides the line joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) in the ratio $-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a{x}_2+b{y}_2+c{z}_2+d}$.

Answer 1

Let:
A = (x₁, y₁, z₁)
B = (x₂, y₂, z₂)

Now, let the line joining A and B be divided by the plane ax + by + cz + d = 0 at point P in the ratio $\lambda :1$.

∴ P = $\left(\frac{\lambda x_2+x_1}{\lambda +1},\frac{\lambda y_2+y_1}{\lambda +1},\frac{\lambda z_2+z_1}{\lambda +1}\right)$
Since P lies on the given plane,

ax + by + cz + d = 0

Thus,

$$\begin{gathered} a\frac{\lambda x_2+x_1}{\lambda +1}+b\frac{\lambda y_2+y_1}{\lambda +1}+c\frac{\lambda z_2+z_1}{\lambda +1}+d=0 \\ \Rightarrow a\left(\lambda x_2+x_1\right)+b\left(\lambda y_2+y_1\right)+c\left(\lambda z_2+z_1\right)+d\left(\lambda +1\right)=0 \\ \Rightarrow \lambda \left(a{x}_2+b{y}_2+c{z}_2+d\right)+\left(a{x}_1+b{y}_1+c{z}_1+d\right)=0 \\ \Rightarrow \lambda \left(a{x}_2+b{y}_2+c{z}_2+d\right)=-\left(a{x}_1+b{y}_1+c{z}_1+d\right) \\ \Rightarrow \lambda =\frac{-\left(a{x}_1+b{y}_1+c{z}_1+d\right)}{\left(a{x}_2+b{y}_2+c{z}_2+d\right)} \\ \Rightarrow \lambda =-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a{x}_2+b{y}_2+c{z}_2+d} \\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{plane}\mathrm{divides}\mathrm{the}\mathrm{line}\mathrm{joining}\left(x_1,y_1,z_1\right)\mathrm{and}\left(x_2,y_2,z_2\right)\mathrm{in}\mathrm{the}\mathrm{ratio}-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a{x}_2+b{y}_2+c{z}_2+d}. \end{gathered}$$