System of Linear Equations: Two planes: angle, intersection & distance — NYJC 2025 H2 Math Prelim Paper 2

(a)

Let \(\theta\) be the acute angle between \(p_1\) and \(p_2\). The normal vectors are \(\mathbf{n}_1=\begin{pmatrix}4\\1\\1\end{pmatrix}\) and \(\mathbf{n}_2=\begin{pmatrix}4\\3\\-1\end{pmatrix}\).

\[\begin{aligned} \cos\theta &= \frac{\left|\mathbf{n}_1\cdot\mathbf{n}_2\right|}{|\mathbf{n}_1||\mathbf{n}_2|} = \frac{|16+3-1|}{\sqrt{18}\cdot\sqrt{26}} = \frac{18}{\sqrt{468}} = \frac{3}{\sqrt{13}} \end{aligned}\]

(b)

\(4(1)+0+4=8\) and \(4(1)+3(0)-4=0\) , so \(A(1,0,4)\) lies on both planes.

The direction of \(l\) is \(\mathbf{n}_1\times\mathbf{n}_2=\begin{pmatrix}4\\1\\1\end{pmatrix}\times\begin{pmatrix}4\\3\\-1\end{pmatrix} =\begin{pmatrix}-4\\8\\8\end{pmatrix}=4\begin{pmatrix}-1\\2\\2\end{pmatrix}\).

(c)

\(\overrightarrow{AB}=\begin{pmatrix}0\\3\\-3\end{pmatrix}\).

\(\overrightarrow{AB}\cdot\begin{pmatrix}-1\\2\\2\end{pmatrix}=0+6-6=0\), so \(AB\) is perpendicular to \(l\).

\(AB=\left|\begin{pmatrix}0\\3\\-3\end{pmatrix}\right|=\sqrt{9+9}=3\sqrt{2}\).

By (a), the perpendicular distance from \(B\) to \(p_2\) is:

\[\begin{aligned} d &= AB\sin\theta\\ &= 3\sqrt{2}\cdot\frac{2}{\sqrt{13}} \end{aligned}\]

(d)

Since \(p_3\) is perpendicular to both \(p_1\) and \(p_2\), the normal vectors of \(p_1\) and \(p_2\) are both parallel to \(p_3\). From (b), a normal vector of \(p_3\) is \(\begin{pmatrix}-1\\2\\2\end{pmatrix}\).

Equation of \(p_3\): \[ \mathbf{r}\cdot\begin{pmatrix}-1\\2\\2\end{pmatrix}=\begin{pmatrix}1\\3\\1\end{pmatrix}\cdot\begin{pmatrix}-1\\2\\2\end{pmatrix}=-1+6+2=7. \]