Scalar product and perpendicular vectorsAQA A-Level Further Maths: Revision notes

Section 1

The scalar product

The scalar product (dot product) of $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$ is a number: $\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta,$ where $\theta$ is the angle between the vectors when both are drawn out of the same point. Example: $\mathbf a=(2,-1,3)$ , $\mathbf b=(4,5,1)$ give $\mathbf a\cdot\mathbf b=8-5+3=6$ . Useful rules: $\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$ , $\mathbf a\cdot(\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c$ , and $\mathbf a\cdot\mathbf a=|\mathbf a|^2$ .

Key termsscalar productmagnitude

Common mistake

Multiplying the components but not adding, which leaves a vector. The scalar product is a single number.

Exam tip

Keep every sign in the products: $(-1)(5)=-5$ .

Section 2

The angle between two vectors

Rearranging the two forms of the scalar product: $\cos\theta=\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}.$ Example: $\mathbf a=(2,-1,3)$ and $\mathbf b=(4,5,1)$ have $|\mathbf a|=\sqrt{14}$ , $|\mathbf b|=\sqrt{42}$ and $\mathbf a\cdot\mathbf b=6$ , so $\cos\theta=\frac{6}{\sqrt{588}}$ and $\theta=75.7^\circ$ . If $\mathbf a\cdot\mathbf b>0$ the angle is acute; if $\mathbf a\cdot\mathbf b<0$ it is obtuse.

Key termsangle between vectors

Common mistake

Forgetting to divide by both magnitudes, or taking $\cos^{-1}$ of the scalar product itself.

Section 3

Perpendicular vectors

For non-zero vectors, the angle is $90^\circ$ exactly when $\cos\theta=0$ , that is $\mathbf a\cdot\mathbf b=0\iff\mathbf a\perp\mathbf b.$ Example: $(2,-1,3)\cdot(1,2,0)=2-2+0=0$ , so they are perpendicular. To find an unknown, set the scalar product to zero. If $(3,k,-2)\perp(2,-4,1)$ then $6-4k-2=0$ , so $k=1$ . In a triangle, a zero scalar product of two sides meeting at a vertex shows a right angle at that vertex.

Key termsperpendicular

Exam tip

Use two sides that start at the same vertex, such as $\overrightarrow{PQ}\cdot\overrightarrow{PR}$ , to test the angle at $P$ .

Section 4

The angle between two lines

The angle between two lines is found from their direction vectors $\mathbf d_1$ and $\mathbf d_2$ ; the base points play no part. $\cos\theta=\frac{|\mathbf d_1\cdot\mathbf d_2|}{|\mathbf d_1||\mathbf d_2|}$ gives the acute angle between the lines. Using the modulus matters because a line has two opposite directions, so the vectors may make an obtuse angle even though the lines make an acute one. Example: $\mathbf d_1=(2,1,-2)$ , $\mathbf d_2=(1,4,8)$ : $\mathbf d_1\cdot\mathbf d_2=-10$ , $|\mathbf d_1|=3$ , $|\mathbf d_2|=9$ , so $\cos\theta=\frac{10}{27}$ and $\theta=68.3^\circ$ . Lines are perpendicular when $\mathbf d_1\cdot\mathbf d_2=0$ , whether or not they meet.

Key termsdirection vectoracute angle

Common mistake

Quoting an obtuse angle ( $111.7^\circ$ ) for the angle between lines. Take the acute one, $180^\circ-111.7^\circ=68.3^\circ$ .

Exam tip

Cartesian form: read the direction from the denominators; vector form: from the multiple of the parameter.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Scalar product and perpendicular vectors

The vectors $\mathbf a=\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$ and $\mathbf b=\begin{pmatrix} 4 \\ 5 \\ 1 \end{pmatrix}$ are given.
Find the angle between $\mathbf a$ and $\mathbf b$ , giving your answer to the nearest $0.1^\circ$ .2 marks
Triangle $PQR$ has vertices $P(1,2,3)$ , $Q(4,0,5)$ and $R(3,4,2)$ .
Hence find the exact area of triangle $PQR$ .2 marks
The lines $l_1$ and $l_2$ have equations $\mathbf r=\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}+\lambda\begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$ and $\mathbf r=\begin{pmatrix} 3 \\ -1 \\ 0 \end{pmatrix}+\mu\begin{pmatrix} 1 \\ 4 \\ 8 \end{pmatrix}$ .
Find the acute angle between $l_1$ and $l_2$ , to the nearest $0.1^\circ$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).