I have to find the equation of the plane that passes through $(0, 0, 0), (4, 0, -2), (0, 8, -6)$.
I have done the following:
The equation of the plane is of the form $$ax+by+cz+d=0$$
Since the points $(0, 0, 0), (4, 0, -2), (0, 8, -6)$ are points o fthe plane, we have that $$d=0 \\ 4a-2c=0 \Rightarrow a=\frac{c}{2} \\ 8b-6c=0 \Rightarrow b=\frac{3c}{4}$$
So $$\frac{c}{2}x+\frac{3c}{4}y+cz=0 \Rightarrow \frac{1}{2}x+\frac{3}{4}y+z=0$$
Is it correct??
Is there also an other way to find the equation of the plane??
Maybe using the cross-product??
Best Answer
Hint: first you find the normal to the plane: $N = \begin{pmatrix} 4 \\ 0 \\ -2 \end{pmatrix} \times \begin{pmatrix} 0 \\ 8 \\-6 \end{pmatrix} = (a,b,c)$, then your plane is:
$a(x-0)+b(y-0)+c(z-0) = 0$