ACJC 2025 H2 Maths Prelim Paper 1 - Vectors Worked Solutions

(a) (b-a)·n=0 so AB⊥n; hence AB is parallel to plane p. (b) r=λ(b-a), λ∈R. (c) OF=c-((c·n)/|n|^2)n.

Free practice / JC2 H2 Maths Prelim

Vectors: Plane normals, lines in a plane & foot of perpendicular — ACJC 2025 H2 Maths Prelim Paper 1

Attempt the question on paper, then open the worked solution to check your method.

Anglo-Chinese Junior College2025 PrelimPaper 1●●○ Standard6 marks

What this question tests

Using the dot product to prove perpendicularity to a plane's normal, writing the vector equation of a line lying in a plane through the origin, and finding the foot of perpendicular from a point to a plane in general vector form.

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 \(\mathbf{n}\).

Show full worked solution▾

(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 \(p\).

Common error: Being perpendicular to the normal makes \(\overrightarrow{AB}\) parallel to \(p\); it does not by itself mean \(\overrightarrow{AB}\) lies inside \(p\) unless a point on \(\overrightarrow{AB}\) is also on \(p\).

(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 \(p\):

\[ \mathbf{r}=\lambda(\mathbf{b}-\mathbf{a}),\qquad \lambda\in\mathbb{R}. \]

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

\[ \overrightarrow{OF}=\mathbf{c}-\dfrac{\mathbf{c}\cdot\mathbf{n}}{|\mathbf{n}|^2}\,\mathbf{n}. \]

Answers: (a) \(\overrightarrow{AB}\cdot\mathbf{n}=0\), so \(\overrightarrow{AB}\perp\mathbf{n}\); hence \(\overrightarrow{AB}\) is parallel to the plane \(p\). (b) \(\mathbf{r}=\lambda(\mathbf{b}-\mathbf{a})\), \(\lambda\in\mathbb{R}\). (c) \(\overrightarrow{OF}=\mathbf{c}-\dfrac{\mathbf{c}\cdot\mathbf{n}}{|\mathbf{n}|^2}\,\mathbf{n}\).

What this question tests

Question

Related questions