Find the equation of the plane with line drawn perpendicular to the planes

Question

$PA$ and $PB$ are drawn perpendicular from $P(α, β, γ)$ to the coordinate planes $x = 0$ and $y = 0$. Show that the equation of the plane OAB is $$\frac{x}α+\frac{y}β-\frac{z}γ=0$$

Answer 1

Note that:

$OA(0, β, γ)$ and $OB(α, 0, γ)$

thus a normal vector is given by the cross product

$$\vec n=OA\times OB=(βγ,αγ,-βα)$$

Thus the plane equation is:

$$βγx+αγy-βαz=0\iff \frac{x}α+\frac{y}β-\frac{z}γ=0$$

Answer 2

Working in homogeneous coordinates, we have $O=[0:0:0:1]$, $A=[0:\beta:\gamma:1]$ and $B=[\alpha:0:\gamma:1]$. The plane through these three points is a homogeneous vector $\mathbf\pi$ such that $O^T\mathbf\pi=A^T\mathbf\pi=B^T\mathbf\pi=0$, i.e., the null vector of the matrix $$\begin{bmatrix}B^T \\ A^T \\ O^T\end{bmatrix} = \begin{bmatrix}\alpha&0&\gamma&1 \\ 0&\beta&\gamma&1 \\ 0&0&0&1\end{bmatrix}.$$ The three points can be taken in any order when forming this matrix, but this one is convenient because the matrix is then already in echelon form. Its null vector can be found by inspection: $$\left[\frac\gamma\alpha:\frac\gamma\beta:-1:0\right] = \left[\frac1\alpha:\frac1\beta:-\frac1\gamma:0\right],$$ which corresponds to the equation $\frac x\alpha+\frac y\beta-\frac z\gamma=0$. We of course require that none of $\alpha$, $\beta$ or $\gamma$ vanish.