Question

Prove that the planes determined by the equations $ax+by+cz=d$ and $\alpha x+\beta y+\gamma z=\delta$ are perpendicular if and only if $\alpha x+\beta y+\gamma z=\delta$

Short answer

The planes determined by the equations $ax+by+cz=d$and $\alpha x+\beta y+\gamma z=\delta$are perpendicular if and only if $a\alpha +b\beta +c\gamma =0$

Step-by-step solution

Given information

The planes determined by the equations $ax+by+cz=d$ and $\alpha x+\beta y+\gamma z=\delta$

Calculation

The goal is to show that planes deduced from provided equations are perpendicular if and only if $a\alpha +b\beta +c\gamma =0$

The normal vector of the equation $ax+by+cz=d$is ${\mathbf{N}}_1=⟨a,b,c⟩$and of the equation $\alpha x+\beta y+\gamma z=\delta$is ${\mathbf{N}}_2=⟨\alpha ,\beta ,\gamma ⟩$

Calculation

Only if and only if the planes' normal vectors are orthogonal are they considered orthogonal.

The normal vectors are orthogonal if and only if ${\mathbf{N}}_1·{\mathbf{N}}_2=0$

The orthogonal condition ${\mathbf{N}}_1·{\mathbf{N}}_2=0$gives:

$$\begin{gathered} ⟨a,b,c⟩·⟨\alpha ,\beta ,\gamma ⟩=0 \\ a\alpha +b\beta +c\gamma =0\text{(Dot Product)} \end{gathered}$$

Thus, the planes determined by the equations $ax+by+cz=d$and $\alpha x+\beta y+\gamma z=\delta$are perpendicular if and only if $a\alpha +b\beta +c\gamma =0$