Problem Statement
Consider the points $A(1, 2, -1)$, $B(2, 0, 3)$, and $C(-1, 1, 2)$.
(a) Find the vector cross product $\vec{AB} \times \vec{AC}$. [4]
(b) Hence, find the Cartesian equation of the plane $\Pi$ passing through $A, B$, and $C$. [3]
Verified Solution & Marking Scheme
(a) Determine displacement vectors AB and AC
$\vec{AB} = \begin{pmatrix} 2 - 1 \\ 0 - 2 \\ 3 - (-1) \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 4 \end{pmatrix}$
$\vec{AC} = \begin{pmatrix} -1 - 1 \\ 1 - 2 \\ 2 - (-1) \end{pmatrix} = \begin{pmatrix} -2 \\ -1 \\ 3 \end{pmatrix}$
Compute cross product AB × AC
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 4 \\ -2 & -1 & 3 \end{vmatrix}$
$= \hat{i}((-2)(3) - (4)(-1)) - \hat{j}((1)(3) - (4)(-2)) + \hat{k}((1)(-1) - (-2)(-2))$
$= \hat{i}(-6 + 4) - \hat{j}(3 + 8) + \hat{k}(-1 - 4) = -2\hat{i} - 11\hat{j} - 5\hat{k} = \begin{pmatrix} -2 \\ -11 \\ -5 \end{pmatrix}$
(b) Form Cartesian equation of plane
Normal vector to the plane $\vec{n} = \begin{pmatrix} 2 \\ 11 \\ 5 \end{pmatrix}$.
Equation: $2x + 11y + 5z = d$.
Substitute coordinates of point $A(1, 2, -1)$:
$d = 2(1) + 11(2) + 5(-1) = 2 + 22 - 5 = 19$
Cartesian equation:
$2x + 11y + 5z = 19$