Problem Statement
Points $A(1, 0, 2)$, $B(2, -1, 3)$, and $C(0, 2, 5)$ define a plane $\Pi$.
**(a)** [3 marks] Find the vector cross product $\vec{AB} \times \vec{AC}$.
**(b)** [2 marks] Hence, write down the Cartesian equation of plane $\Pi$.
**(c)** [3 marks] Find the shortest perpendicular distance from the point $D(3, 1, 4)$ to the plane $\Pi$.
Verified Solution & Marking Scheme
Part (a) [M1]
$\vec{AB} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$, $\vec{AC} = \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}$
Part (a) [A1]
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ -1 & 2 & 3 \end{vmatrix} = \hat{i}(-3 - 2) - \hat{j}(3 - (-1)) + \hat{k}(2 - 1)$
Part (a) [A1]
$\vec{n} = \begin{pmatrix} -5 \\ -4 \\ 1 \end{pmatrix}$ (or $\begin{pmatrix} 5 \\ 4 \\ -1 \end{pmatrix}$)
Part (b) [M1]
Equation $5x + 4y - z = d$. Using point $A(1, 0, 2)$: $5(1) + 4(0) - (2) = 3$
Part (b) [A1]
$5x + 4y - z = 3$ (or $-5x - 4y + z = -3$)
Part (c) [M1]
Formula $d = \frac{|a x_0 + b y_0 + c z_0 - d|}{\sqrt{a^2 + b^2 + c^2}}$
Part (c) [A1]
$d = \frac{|5(3) + 4(1) - 4 - 3|}{\sqrt{5^2 + 4^2 + (-1)^2}} = \frac{|15 + 4 - 4 - 3|}{\sqrt{25 + 16 + 1}} = \frac{12}{\sqrt{42}}$
Part (c) [A1]
$d = \frac{2\sqrt{42}}{7}\text{ units} \approx 1.85$