Question

Find the Cartesian equation of the plane passing through the points $A(2,5,-3)$, $B(-2,-3,5)$ and $C(5,3,-3)$.

Answer 1

Three points (A,B,C) can define two distinct vectors AB and AC. Since the two vectors lie on the plane, their cross product can be used as a normal to the plane.

• Determine the vectors
  
• Find the cross product of the two vectors
• Substitute one point into the Cartesian equation to solve for d.

$\overrightarrow{AB}=(x_B-x_A)\hat{i}+(y_B-y_A)\hat{j}+(z_B-z_A)\hat{k}$$⟹\overrightarrow{AB}=-4\hat{i}-8\hat{j}+8\hat{k}\overrightarrow{AC}=(x_c-x_A)\hat{i}+(y_c-y_A)\hat{j}+(z_c-z_A)\hat{k}⟹\overrightarrow{AC}=3\hat{i}-2\hat{j}\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -4& -8& 8\\ 3& -2& 0\end{array}\right|=16\hat{i}+24\hat{j}+32\hat{k}$
The equation of the plane is $16x+24y+32z=d$
Plug any point in the equation to fing $d$

$⟹16\left(2\right)+24\left(5\right)+32(-3)=d⟹56=d$

The equation of the plane is $16x+24y+32z=56$