Mathematics 9709 · Pure Mathematics 3

The scalar (dot) product

The scalar product is the tool that turns vectors into angles.
It has two forms: $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3$ (multiply matching components, then add) and $|\mathbf{a}||\mathbf{b}|\cos\theta$ (the lengths times the cosine of the angle). Both give the same number. Set them equal and you get the angle.
If the dot comes out zero, the vectors are at right angles.

The angle. $\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ : dot the vectors, divide by the two lengths multiplied together, then take $\cos^{-1}$ .

The right angle. $\mathbf{a}\cdot\mathbf{b}=0$ is the right-angle test. No lengths, no angle, just “does the dot come out zero?”. Every “show these are perpendicular”, rectangle or right-angle mark is this one line.

Two formulas, one number

$\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3=|\mathbf{a}||\mathbf{b}|\cos\theta$

The angle θ between a and b sits at their common tail; cos θ = (a·b)/(|a||b|).

Drag $\mathbf{b}$ below and watch the dot and the angle $\theta$ move together; when the dot hits zero, the angle is $90^\circ$ .

b acrossb up

a = (3, 1)·b = (1, 3)·a·b = 6·θ = 53.1°

Why a zero dot product means a right angle

Look at the second form: $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ . The lengths $|\mathbf{a}|$ and $|\mathbf{b}|$ are always positive, so the dot product has the same sign as $\cos\theta$ . And $\cos\theta=0$ at only one angle between 0 and 180 degrees: $\theta=90^\circ$ . So “dot product is zero” and “perpendicular” say the same thing.

Worked examples

Worked example

Real question: show $OABC$ is a rectangle, $\overrightarrow{OA}=5\mathbf{i}-2\mathbf{j}+\mathbf{k}$ , $\overrightarrow{OB}=8\mathbf{i}+2\mathbf{j}-6\mathbf{k}$ , $\overrightarrow{OC}=3\mathbf{i}+4\mathbf{j}-7\mathbf{k}$

9709/32 F/M 2024 Q9(a). A rectangle needs a right angle at the corner, and one dot product settles it.

Adjacent sides at O are $\overrightarrow{OA}$ and $\overrightarrow{OC}$ . Dot them: $\overrightarrow{OA}\cdot\overrightarrow{OC}=(5)(3)+(-2)(4)+(1)(-7)=15-8-7=0$ .
Zero, so the corner at O is a right angle. (The full question then checks $\overrightarrow{AB}=\overrightarrow{OC}$ to confirm the opposite sides match.)

Answer

\overrightarrow{OA}\cdot\overrightarrow{OC}=0\ \Rightarrow\ \text{right angle at }O

Worked example

Real question: $A(1,3,1)$ , $B(3,1,2)$ , $C(5,3,-2)$ : find $\cos$ of angle $BAC$

9709/32 O/N 2022 Q6(a). For an angle at a corner, build both arrows out of that corner A, then dot them.

The angle sits at $A$ , so make both steps start at $A$ : $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(2,-2,1)$ , $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=(4,0,-3)$ .
Dot them: $\overrightarrow{AB}\cdot\overrightarrow{AC}=(2)(4)+(-2)(0)+(1)(-3)=8-3=5$ .
Lengths: $|\overrightarrow{AB}|=\sqrt{4+4+1}=3$ , $|\overrightarrow{AC}|=\sqrt{16+9}=5$ .
Divide: $\cos(\angle BAC)=\dfrac{5}{3\times5}=\dfrac{1}{3}$ .

Answer

\cos(\angle BAC)=\tfrac{1}{3}

Worked example

Real question: $A(2,0,5)$ , $B(3,2,3)$ , $C(1,1,1)$ : show angle $ABC$ is a right angle, and $ABC$ isosceles

9709/31 M/J 2020 Q9(a),(b). The angle is at $B$ , so both arrows must start at $B$ .

Build both steps from $B$ : $\overrightarrow{BA}=\mathbf{a}-\mathbf{b}=(-1,-2,2)$ , $\overrightarrow{BC}=\mathbf{c}-\mathbf{b}=(-2,-1,-2)$ .
Dot: $\overrightarrow{BA}\cdot\overrightarrow{BC}=(-1)(-2)+(-2)(-1)+(2)(-2)=2+2-4=0$ , so the angle at $B$ is $90^\circ$ .
Isosceles? Compare the two sides at $B$ : $|\overrightarrow{BA}|=\sqrt{1+4+4}=3$ and $|\overrightarrow{BC}|=\sqrt{4+1+4}=3$ . Equal, so two sides match.

Answer

\overrightarrow{BA}\cdot\overrightarrow{BC}=0\ \Rightarrow\ \angle ABC=90^\circ;\ \ |BA|=|BC|=3\ \Rightarrow\ \text{isosceles}

Worked example

Angle between $\mathbf{p}=(8,-6,-10)$ and $\mathbf{q}=(4,-3,0)$

Dot product: $\mathbf{p}\cdot\mathbf{q}=(8)(4)+(-6)(-3)+(-10)(0)=32+18+0=50$ .
Magnitudes: $|\mathbf{p}|=\sqrt{64+36+100}=\sqrt{200}=10\sqrt2$ , $|\mathbf{q}|=\sqrt{16+9}=5$ .
Divide: $\cos\theta=\dfrac{50}{10\sqrt2\cdot5}=\dfrac{1}{\sqrt2}$ , so $\theta=45^\circ$ .

Answer

\theta=45^\circ

Worked example

Find $a$ so that $(6,-2,5)$ and $(a,4,-2)$ are perpendicular

Perpendicular means the dot product is zero: $(6)(a)+(-2)(4)+(5)(-2)=0$ , i.e. $6a-8-10=0$ .
Solve: $6a=18$ .

Answer

a=3

Where the marks go

Common mistake

The lengths go on the bottom:

\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

, not

\cos\theta=\mathbf{a}\cdot\mathbf{b}

. And in 3D keep all three components when you square for

|\mathbf{a}|

Common mistake

When the question is about a shape (a rectangle, the diagonals of OABC), build the right vectors first. A side is one position vector minus another, and a diagonal joins opposite corners. Dot the wrong pair and you lose the method mark even if the arithmetic is perfect. (9709/32 Feb/March 2024 Q9)

Common mistake

For an angle at a corner like

\angle BAC

, both arrows must start at that corner:

\overrightarrow{AB}

and

\overrightarrow{AC}

, not

\overrightarrow{BA}

and not raw position vectors. Dotting

\overrightarrow{AB}

with

\overrightarrow{CA}

flips the sign and gives you the wrong angle. The examiner only scores

\overrightarrow{AB}\cdot\overrightarrow{AC}

. (9709/32 Oct/Nov 2022 Q6)

Now you try

Are $\mathbf{a}=(2,8,-2)$ and $\mathbf{b}=(7,-1,3)$ at right angles? (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\mathbf{a}\cdot\mathbf{b}=(2)(7)+(8)(-1)+(-2)(3)=14-8-6=0$ , and zero means a right angle:

Answer

\text{yes, perpendicular}

PreviousUnit vectors NextThe vector equation of a line

The scalar (dot) product

The scalar product is the tool that turns vectors into angles.
It has two forms: $\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3$ (multiply matching components, then add) and $|\mathbf{a}||\mathbf{b}|\cos\theta$ (the lengths times the cosine of the angle). Both give the same number. Set them equal and you get the angle.
If the dot comes out zero, the vectors are at right angles.

The angle. $\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ : dot the vectors, divide by the two lengths multiplied together, then take $\cos^{-1}$ .

Two formulas, one number

$\mathbf{a}\cdot\mathbf{b}=a_1b_1+a_2b_2+a_3b_3=|\mathbf{a}||\mathbf{b}|\cos\theta$

The angle θ between a and b sits at their common tail; cos θ = (a·b)/(|a||b|).

Drag $\mathbf{b}$ below and watch the dot and the angle $\theta$ move together; when the dot hits zero, the angle is $90^\circ$ .

b acrossb up

a = (3, 1)·b = (1, 3)·a·b = 6·θ = 53.1°

Why a zero dot product means a right angle

Worked examples

Worked example

9709/32 F/M 2024 Q9(a). A rectangle needs a right angle at the corner, and one dot product settles it.

Adjacent sides at O are $\overrightarrow{OA}$ and $\overrightarrow{OC}$ . Dot them: $\overrightarrow{OA}\cdot\overrightarrow{OC}=(5)(3)+(-2)(4)+(1)(-7)=15-8-7=0$ .
Zero, so the corner at O is a right angle. (The full question then checks $\overrightarrow{AB}=\overrightarrow{OC}$ to confirm the opposite sides match.)

Answer

\overrightarrow{OA}\cdot\overrightarrow{OC}=0\ \Rightarrow\ \text{right angle at }O

Worked example

Real question: $A(1,3,1)$ , $B(3,1,2)$ , $C(5,3,-2)$ : find $\cos$ of angle $BAC$

9709/32 O/N 2022 Q6(a). For an angle at a corner, build both arrows out of that corner A, then dot them.

The angle sits at $A$ , so make both steps start at $A$ : $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=(2,-2,1)$ , $\overrightarrow{AC}=\mathbf{c}-\mathbf{a}=(4,0,-3)$ .
Dot them: $\overrightarrow{AB}\cdot\overrightarrow{AC}=(2)(4)+(-2)(0)+(1)(-3)=8-3=5$ .
Lengths: $|\overrightarrow{AB}|=\sqrt{4+4+1}=3$ , $|\overrightarrow{AC}|=\sqrt{16+9}=5$ .
Divide: $\cos(\angle BAC)=\dfrac{5}{3\times5}=\dfrac{1}{3}$ .

Answer

\cos(\angle BAC)=\tfrac{1}{3}

Worked example

Real question: $A(2,0,5)$ , $B(3,2,3)$ , $C(1,1,1)$ : show angle $ABC$ is a right angle, and $ABC$ isosceles

9709/31 M/J 2020 Q9(a),(b). The angle is at $B$ , so both arrows must start at $B$ .

Build both steps from $B$ : $\overrightarrow{BA}=\mathbf{a}-\mathbf{b}=(-1,-2,2)$ , $\overrightarrow{BC}=\mathbf{c}-\mathbf{b}=(-2,-1,-2)$ .
Dot: $\overrightarrow{BA}\cdot\overrightarrow{BC}=(-1)(-2)+(-2)(-1)+(2)(-2)=2+2-4=0$ , so the angle at $B$ is $90^\circ$ .
Isosceles? Compare the two sides at $B$ : $|\overrightarrow{BA}|=\sqrt{1+4+4}=3$ and $|\overrightarrow{BC}|=\sqrt{4+1+4}=3$ . Equal, so two sides match.

Answer

\overrightarrow{BA}\cdot\overrightarrow{BC}=0\ \Rightarrow\ \angle ABC=90^\circ;\ \ |BA|=|BC|=3\ \Rightarrow\ \text{isosceles}

Worked example

Angle between $\mathbf{p}=(8,-6,-10)$ and $\mathbf{q}=(4,-3,0)$

Dot product: $\mathbf{p}\cdot\mathbf{q}=(8)(4)+(-6)(-3)+(-10)(0)=32+18+0=50$ .
Magnitudes: $|\mathbf{p}|=\sqrt{64+36+100}=\sqrt{200}=10\sqrt2$ , $|\mathbf{q}|=\sqrt{16+9}=5$ .
Divide: $\cos\theta=\dfrac{50}{10\sqrt2\cdot5}=\dfrac{1}{\sqrt2}$ , so $\theta=45^\circ$ .

Answer

\theta=45^\circ

Worked example

Find $a$ so that $(6,-2,5)$ and $(a,4,-2)$ are perpendicular

Perpendicular means the dot product is zero: $(6)(a)+(-2)(4)+(5)(-2)=0$ , i.e. $6a-8-10=0$ .
Solve: $6a=18$ .

Answer

a=3

Where the marks go

Common mistake

The lengths go on the bottom:

\cos\theta=\dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}

, not

\cos\theta=\mathbf{a}\cdot\mathbf{b}

. And in 3D keep all three components when you square for

|\mathbf{a}|

Common mistake

For an angle at a corner like

\angle BAC

, both arrows must start at that corner:

\overrightarrow{AB}

and

\overrightarrow{AC}

, not

\overrightarrow{BA}

and not raw position vectors. Dotting

\overrightarrow{AB}

with

\overrightarrow{CA}

flips the sign and gives you the wrong angle. The examiner only scores

\overrightarrow{AB}\cdot\overrightarrow{AC}

. (9709/32 Oct/Nov 2022 Q6)

Now you try

Are $\mathbf{a}=(2,8,-2)$ and $\mathbf{b}=(7,-1,3)$ at right angles? (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\mathbf{a}\cdot\mathbf{b}=(2)(7)+(8)(-1)+(-2)(3)=14-8-6=0$ , and zero means a right angle:

Answer

\text{yes, perpendicular}