Question

Determine the equation of the plane containing the points P(1,3,2) Q=(-2,0,-2) and R=(1,4,3)

Answer 1

There points

$\begin{aligned} P=(1,3,2)\\ Q=(-2,0,2) \\ R=(1,4,3) \end{aligned}$

So, the vector$\overrightarrow {PQ}$ and $\overrightarrow{PR}$ will be

$\begin{aligned} \overrightarrow{PQ} =(-2-1) \hat{\imath}+(0-3) \hat{\jmath}+(2-2) \hat{k}=-3 \hat{i}-3 \hat{\jmath} \\ \overrightarrow{PR}=(1-1) \hat{i}+(4-3) \hat{\jmath}+(3-2) \hat{k}=\hat{\jmath}+\hat{k} \\ \end{aligned}$

So, The vector perpendicular to this two vectors is given by

$\overline{P Q} \times \overline{P R}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -3 & -3 & 0 \\ 0 & 1 & 1 \end{array}\right|$

So. The equation of plane will be

$\begin{gathered} -3(x-1)+3(y-3)+(-3)(z-2)=0 \\ \Rightarrow x-y+z=2 \end{gathered}$