Mathematics 9709 · Pure Mathematics 3

The angle between two lines

The angle between two lines is just the angle between their direction vectors. Only which way they point matters, not where they sit.
So it is the scalar product from §9.3, with the two directions fed in.
One thing to watch: put the modulus on top so you always get the acute angle the question wants.

$\cos\theta=\dfrac{|\mathbf{d_1}\cdot\mathbf{d_2}|}{|\mathbf{d_1}|\,|\mathbf{d_2}|}$ : take the directions straight from the brackets, dot them, divide by the lengths. The bars force an acute answer.

Only the directions matter. The start points $\mathbf{a}$ don't come into it. So this even works for skew lines that never meet: there is still a clear angle between the ways they point.

Scalar product of the directions (mod on top for acute)

$\cos\theta=\dfrac{|\mathbf{d_1}\cdot\mathbf{d_2}|}{|\mathbf{d_1}|\,|\mathbf{d_2}|}$

Take $\mathbf{d_1}$ and $\mathbf{d_2}$ straight from the $s$ and $t$ brackets, where the bars keep $\cos\theta\ge0$ so the angle stays acute.

Slide both directions to the same tail: θ between them is the angle between the lines.

Why the modulus gives the acute angle

Flip a direction vector round (use $-\mathbf{d_1}$ instead of $\mathbf{d_1}$ ) and the line is the same, but the dot product changes sign. That turns the angle from $\theta$ into $180^\circ-\theta$ . Both count as “the angle between the lines”, so the question makes the answer definite by asking for the acute one. Taking $|\mathbf{d_1}\cdot\mathbf{d_2}|$ keeps the cosine positive, which forces $\theta\le90^\circ$ every time.

Worked examples

Worked example

Real question: acute angle between lines with directions $(2,-1,3)$ and $(1,-1,4)$

9709/32 F/M 2021 Q7(b). These two lines turn out skew in §9.5, but the angle doesn't care whether they meet.

Dot the directions: $\mathbf{d_1}\cdot\mathbf{d_2}=(2)(1)+(-1)(-1)+(3)(4)=2+1+12=15$ .
Lengths: $|\mathbf{d_1}|=\sqrt{4+1+9}=\sqrt{14}$ , $|\mathbf{d_2}|=\sqrt{1+1+16}=\sqrt{18}$ .
Divide: $\cos\theta=\dfrac{15}{\sqrt{14}\,\sqrt{18}}=\dfrac{15}{\sqrt{252}}=0.9449\ldots$
Inverse cosine: $\theta=\cos^{-1}(0.9449\ldots)$ .

Answer

\theta=19.1^\circ

Worked example

Real question: acute angle between directions $(-1,-3,1)$ and $(1,-3,-2)$

9709/32 F/M 2022 Q10(b): the angle between line $AB$ from §9.4 and the given line $l$ . The dot is positive here, so the modulus changes nothing, but keep the bars on anyway.

Dot the directions: $(-1)(1)+(-3)(-3)+(1)(-2)=-1+9-2=6$ .
Lengths: $\sqrt{1+9+1}=\sqrt{11}$ , $\sqrt{1+9+4}=\sqrt{14}$ .
Divide: $\cos\theta=\dfrac{6}{\sqrt{11}\,\sqrt{14}}=\dfrac{6}{\sqrt{154}}=0.4834\ldots$
Inverse cosine: $\theta=\cos^{-1}(0.4834\ldots)$ .

Answer

\theta=61.1^\circ

Where the marks go

Common mistake

Use the directions, not the start points

\mathbf{a}

. The points only say where the lines are. The angle depends on which way they point.

Common mistake

If a negative dot product gives you an obtuse angle, the bars

|\mathbf{d_1}\cdot\mathbf{d_2}|

fix it. An “acute angle” question wants an answer below

90^\circ

, so if you forgot the modulus, take your angle away from

180^\circ

. (9709/32 Feb/March 2021 Q7)

Now you try

Find the acute angle between the directions $\mathbf{d_1}=(1,-8,5)$ and $\mathbf{d_2}=(5,-1,2)$ . (9709-style)

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

$\mathbf{d_1}\cdot\mathbf{d_2}=(1)(5)+(-8)(-1)+(5)(2)=23$ , $|\mathbf{d_1}|=\sqrt{90}$ , $|\mathbf{d_2}|=\sqrt{30}$ :

$\cos\theta=\dfrac{23}{\sqrt{90}\,\sqrt{30}}=0.4426\ldots$ , so

Answer

\theta=63.7^\circ

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The angle between two lines

The angle between two lines is just the angle between their direction vectors. Only which way they point matters, not where they sit.
So it is the scalar product from §9.3, with the two directions fed in.
One thing to watch: put the modulus on top so you always get the acute angle the question wants.

$\cos\theta=\dfrac{|\mathbf{d_1}\cdot\mathbf{d_2}|}{|\mathbf{d_1}|\,|\mathbf{d_2}|}$ : take the directions straight from the brackets, dot them, divide by the lengths. The bars force an acute answer.

Only the directions matter. The start points $\mathbf{a}$ don't come into it. So this even works for skew lines that never meet: there is still a clear angle between the ways they point.

Scalar product of the directions (mod on top for acute)

$\cos\theta=\dfrac{|\mathbf{d_1}\cdot\mathbf{d_2}|}{|\mathbf{d_1}|\,|\mathbf{d_2}|}$

Take $\mathbf{d_1}$ and $\mathbf{d_2}$ straight from the $s$ and $t$ brackets, where the bars keep $\cos\theta\ge0$ so the angle stays acute.

Slide both directions to the same tail: θ between them is the angle between the lines.

Why the modulus gives the acute angle

Worked examples

Worked example

Real question: acute angle between lines with directions $(2,-1,3)$ and $(1,-1,4)$

9709/32 F/M 2021 Q7(b). These two lines turn out skew in §9.5, but the angle doesn't care whether they meet.

Dot the directions: $\mathbf{d_1}\cdot\mathbf{d_2}=(2)(1)+(-1)(-1)+(3)(4)=2+1+12=15$ .
Lengths: $|\mathbf{d_1}|=\sqrt{4+1+9}=\sqrt{14}$ , $|\mathbf{d_2}|=\sqrt{1+1+16}=\sqrt{18}$ .
Divide: $\cos\theta=\dfrac{15}{\sqrt{14}\,\sqrt{18}}=\dfrac{15}{\sqrt{252}}=0.9449\ldots$
Inverse cosine: $\theta=\cos^{-1}(0.9449\ldots)$ .

Answer

\theta=19.1^\circ

Worked example

Real question: acute angle between directions $(-1,-3,1)$ and $(1,-3,-2)$

9709/32 F/M 2022 Q10(b): the angle between line $AB$ from §9.4 and the given line $l$ . The dot is positive here, so the modulus changes nothing, but keep the bars on anyway.

Dot the directions: $(-1)(1)+(-3)(-3)+(1)(-2)=-1+9-2=6$ .
Lengths: $\sqrt{1+9+1}=\sqrt{11}$ , $\sqrt{1+9+4}=\sqrt{14}$ .
Divide: $\cos\theta=\dfrac{6}{\sqrt{11}\,\sqrt{14}}=\dfrac{6}{\sqrt{154}}=0.4834\ldots$
Inverse cosine: $\theta=\cos^{-1}(0.4834\ldots)$ .

Answer

\theta=61.1^\circ

Where the marks go

Common mistake

Use the directions, not the start points

\mathbf{a}

. The points only say where the lines are. The angle depends on which way they point.

Common mistake

If a negative dot product gives you an obtuse angle, the bars

|\mathbf{d_1}\cdot\mathbf{d_2}|

fix it. An “acute angle” question wants an answer below

90^\circ

, so if you forgot the modulus, take your angle away from

180^\circ

. (9709/32 Feb/March 2021 Q7)

Now you try

Find the acute angle between the directions $\mathbf{d_1}=(1,-8,5)$ and $\mathbf{d_2}=(5,-1,2)$ . (9709-style)

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

$\mathbf{d_1}\cdot\mathbf{d_2}=(1)(5)+(-8)(-1)+(5)(2)=23$ , $|\mathbf{d_1}|=\sqrt{90}$ , $|\mathbf{d_2}|=\sqrt{30}$ :

$\cos\theta=\dfrac{23}{\sqrt{90}\,\sqrt{30}}=0.4426\ldots$ , so

Answer

\theta=63.7^\circ