2024 H2 Maths Paper 1 Q2 — Showing Two Vectors Are Perpendicular

The two vectors, livethe parallelogram

The angle between them,

\theta

35°

The unit vector

\mathbf{a}

The vector

\mathbf{b}

, length 3 The parallelogram, area

\left|\mathbf{a}\times\mathbf{b}\right|

Showing $\mathbf{a}$ and $\mathbf{b}$ are perpendicular

3 marks

Two vectors $\mathbf{a}$ and $\mathbf{b}$ satisfy $\left|\mathbf{a}\times\mathbf{b}\right|=3$ . Given that $\mathbf{a}$ is a unit vector and $\mathbf{b}\cdot\mathbf{b}=9$ , show that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular.

Approach

Everything here is a length in disguise. "Unit vector" gives $\left|\mathbf{a}\right|=1$ ; $\mathbf{b}\cdot\mathbf{b}=9$ gives $\left|\mathbf{b}\right|=3$ . Feed both into $\left|\mathbf{a}\times\mathbf{b}\right|=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin\theta$ and the angle has nowhere to hide.

Since $\mathbf{a}$ is a unit vector,

\left|\mathbf{a}\right|=1

and since a vector dotted with itself gives the square of its length,

\mathbf{b}\cdot\mathbf{b}=\left|\mathbf{b}\right|^{2}=9\ \Longrightarrow\ \left|\mathbf{b}\right|=3

(taking the positive root, as a magnitude cannot be negative). Let $\theta$ be the angle between $\mathbf{a}$ and $\mathbf{b}$ . Then

\left|\mathbf{a}\times\mathbf{b}\right|=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin\theta

3=(1)(3)\sin\theta\ \Longrightarrow\ \sin\theta=1

The angle between two vectors is taken in $0^\circ\leq\theta\leq180^\circ$ , and on that interval $\sin\theta=1$ has the single solution

\theta=90^\circ

To check your answer:

$\left|\mathbf{b}\right|=\sqrt{9}=3$ , and $(1)(3)\sin90^\circ=3$ , which is the value of $\left|\mathbf{a}\times\mathbf{b}\right|$ given ✓. As a sanity check on the logic: $\left|\mathbf{a}\right|\left|\mathbf{b}\right|=3$ is the largest $\left|\mathbf{a}\times\mathbf{b}\right|$ could ever be, and the question hands you exactly that value.

$\therefore\ \theta=90^\circ$ , so $\mathbf{a}$ and $\mathbf{b}$ are perpendicular. (shown)

Where the marks go

Two things earn the third mark. First, $\mathbf{b}\cdot\mathbf{b}=9$ must be converted — writing $\left|\mathbf{b}\right|=9$ loses the question immediately. Second, the range matters: $\sin\theta=1$ only pins $\theta$ down because the angle between two vectors lies in $[0^\circ,180^\circ]$ . Saying "so $\theta=90^\circ$ " without that line is an assertion, not a proof — and this is a show question, where the reasoning is the answer.

Why $\sin\theta=1$ is the lucky case

worth knowing

Not part of the question — but the reason this one works in three lines, while its near-neighbours do not.

Suppose the question had said $\left|\mathbf{a}\times\mathbf{b}\right|=\frac{3}{2}$ instead. Then

\sin\theta=\tfrac{1}{2}\ \Longrightarrow\ \theta=30^\circ\ \text{or}\ \theta=150^\circ

and the angle would not be determined: the sine function takes each value below 1 twice on $[0^\circ,180^\circ]$ , once acute and once obtuse. It is only at the maximum, $\sin\theta=1$ , that the two solutions collapse into one. That is why the question could set the cross product to its largest possible value and still ask you to show something.

Note — the scalar product settles the same thing faster

If a question gives you $\mathbf{a}\cdot\mathbf{b}$ rather than $\left|\mathbf{a}\times\mathbf{b}\right|$ , perpendicularity is immediate: $\mathbf{a}\cdot\mathbf{b}=0$ and you are done, with no ambiguity to discuss, because $\cos\theta$ takes each value exactly once on $[0^\circ,180^\circ]$ . Drag the slider and watch the two readouts: the dot product passes through zero once, while the cross product's magnitude hits each value twice.

What this question was really testing

$\mathbf{b}\cdot\mathbf{b}=\left|\mathbf{b}\right|^{2}$ . A self-dot-product is a squared length. It is the single most common way an examiner hands you a magnitude without saying "magnitude".
The cross product formula both ways. $\left|\mathbf{a}\times\mathbf{b}\right|=\left|\mathbf{a}\right|\left|\mathbf{b}\right|\sin\theta$ is usually used to find an area; here it is used backwards, to find the angle.
Quoting the range. The angle between two vectors is in $[0^\circ,180^\circ]$ . Stating it is what turns $\sin\theta=1$ into a unique angle.
Recognising a maximum. $\left|\mathbf{a}\right|\left|\mathbf{b}\right|$ is the ceiling for $\left|\mathbf{a}\times\mathbf{b}\right|$ . Being told the cross product reaches it is the same as being told the vectors are perpendicular — the algebra just confirms what the geometry already says.

A cross product at full stretch, and the angle that forces

Showing $\mathbf{a}$ and $\mathbf{b}$ are perpendicular

Why $\sin\theta=1$ is the lucky case

What this question was really testing

This is how we mark, and how we teach.

A cross product at full stretch, and the angle that forces

Showing a\mathbf{a} and b\mathbf{b} are perpendicular

Why sin⁡θ=1\sin\theta=1 is the lucky case

What this question was really testing

This is how we mark, and how we teach.

Showing $\mathbf{a}$ and $\mathbf{b}$ are perpendicular

Why $\sin\theta=1$ is the lucky case