Problem Statement
Show that the four points with position vectors $A(4\hat{i} + 5\hat{j} + \hat{k})$, $B(-\hat{j} - \hat{k})$, $C(3\hat{i} + 9\hat{j} + 4\hat{k})$, and $D(-4\hat{i} + 4\hat{j} + 4\hat{k})$ are coplanar.
Verified Solution & Marking Scheme
Find coterminous vectors AB, AC, AD
$\vec{AB} = (0 - 4)\hat{i} + (-1 - 5)\hat{j} + (-1 - 1)\hat{k} = -4\hat{i} - 6\hat{j} - 2\hat{k}$
$\vec{AC} = (3 - 4)\hat{i} + (9 - 5)\hat{j} + (4 - 1)\hat{k} = -\hat{i} + 4\hat{j} + 3\hat{k}$
$\vec{AD} = (-4 - 4)\hat{i} + (4 - 5)\hat{j} + (4 - 1)\hat{k} = -8\hat{i} - \hat{j} + 3\hat{k}$
Evaluate Scalar Triple Product [AB AC AD]
$[\vec{AB} \, \vec{AC} \, \vec{AD}] = \begin{vmatrix} -4 & -6 & -2 \\ -1 & 4 & 3 \\ -8 & -1 & 3 \end{vmatrix}$
$= -4(12 - (-3)) - (-6)(-3 - (-24)) + (-2)(1 - (-32))$
$= -4(15) + 6(21) - 2(33) = -60 + 126 - 66 = 0$
Since the scalar triple product is 0, the vectors $\vec{AB}, \vec{AC}, \vec{AD}$ are coplanar.
Therefore, the points $A, B, C, D$ are coplanar.