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,

\[a\frac{\lambda x_2 + x_1}{\lambda + 1} + b\frac{\lambda y_2 + y_1}{\lambda + 1} + c\frac{\lambda z_2 + z_1}{\lambda + 1} + d = 0\]
\[ \Rightarrow a\left( \lambda x_2 + x_1 \right) + b\left( \lambda y_2 + y_1 \right) + c\left( \lambda z_2 + z_1 \right) + d\left( \lambda + 1 \right) = 0\]
\[ \Rightarrow \lambda\left( a x_2 + b y_2 + c z_2 + d \right) + \left( a x_1 + b y_1 + c z_1 + d \right) = 0\]
\[ \Rightarrow \lambda\left( a x_2 + b y_2 + c z_2 + d \right) = - \left( a x_1 + b y_1 + c z_1 + d \right)\]
\[ \Rightarrow \lambda = \frac{- \left( a x_1 + b y_1 + c z_1 + d \right)}{\left( a x_2 + b y_2 + c z_2 + d \right)}\]
\[ \Rightarrow \lambda = - \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]
\[\text{ Thus, the given plane divides the line joining }  \left( x_1 , y_1 , z_1 \right) and \left( x_2 , y_2 , z_2 \right) in the ratio - \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d} .\]