Vectors: Plane normals, lines in a plane & foot of perpendicular — ACJC 2025 H2 Math Prelim P1
What this question tests
Question
Relative to the origin \(O\), the points \(A\) and \(B\) have non-zero and non-parallel position vectors \(\mathbf{a}\) and \(\mathbf{b}\) respectively. The plane \(p\) has equation \(\mathbf{r}\cdot\mathbf{n}=0\).
(a) Given that \(\mathbf{a}\cdot\mathbf{n}=\mathbf{b}\cdot\mathbf{n}\neq 0\), show that \(\overrightarrow{AB}\) is perpendicular to \(\mathbf{n}\). Hence, describe the geometrical relationship between \(\overrightarrow{AB}\) and the plane \(p\).
(b) Write down the equation of a line parallel to \(\overrightarrow{AB}\) that is contained in the plane \(p\).
(c) The point \(F\) is the foot of perpendicular from a point \(C\) with position vector \(\mathbf{c}\) to the plane \(p\). Find the position vector of point \(F\), giving your answer in terms of \(\mathbf{c}\) and \(\mathbf{n}\).
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(a) Show \(\overrightarrow{AB}\perp\mathbf{n}\) and describe the relationship with \(p\)
Subtract the two equal dot products so that the difference \(\mathbf{b}-\mathbf{a}=\overrightarrow{AB}\) appears.
Since \(\mathbf{a}\cdot\mathbf{n}=\mathbf{b}\cdot\mathbf{n}\),
\[ (\mathbf{b}-\mathbf{a})\cdot\mathbf{n}=0, \]so \(\overrightarrow{AB}\cdot\mathbf{n}=0\), which means \(\overrightarrow{AB}\perp\mathbf{n}\).
Since \(\mathbf{n}\) is the normal to the plane \(p\) and \(\overrightarrow{AB}\) is perpendicular to \(\mathbf{n}\), \(\overrightarrow{AB}\) is parallel to the plane \(p\).
(b) A line parallel to \(\overrightarrow{AB}\) contained in \(p\)
The origin lies on \(p\), so a line through \(O\) in the direction \(\mathbf{b}-\mathbf{a}\) stays in the plane.
Since \(\mathbf{r}=\mathbf{0}\) satisfies \(\mathbf{r}\cdot\mathbf{n}=0\), the origin \(O\) lies on \(p\). A line through \(O\) with direction \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}\) therefore lies in \(p\):
\[ \mathbf{r}=\lambda(\mathbf{b}-\mathbf{a}),\qquad \lambda\in\mathbb{R}. \](c) Position vector of the foot of perpendicular \(F\)
Drop a line from \(C\) along the normal \(\mathbf{n}\) and find where it meets the plane.
The line through \(C\) perpendicular to \(p\) has equation \(\mathbf{r}=\mathbf{c}+\mu\mathbf{n}\). Substituting into \(\mathbf{r}\cdot\mathbf{n}=0\):
\[ \begin{aligned} (\mathbf{c}+\mu\mathbf{n})\cdot\mathbf{n} &= 0\\ \mathbf{c}\cdot\mathbf{n}+\mu|\mathbf{n}|^2 &= 0\\ \mu &= -\dfrac{\mathbf{c}\cdot\mathbf{n}}{|\mathbf{n}|^2}. \end{aligned} \]Hence the position vector of \(F\) is
\[ \overrightarrow{OF}=\mathbf{c}-\dfrac{\mathbf{c}\cdot\mathbf{n}}{|\mathbf{n}|^2}\,\mathbf{n}. \]