Core Pure · Vectors · Year 1

Calculating Angles Between Lines and Planes

Line-line, line-plane, plane-plane — three variations on a theme, distinguished by a single question: is it sine or cosine you need?

17 min read 4 worked examples

Angle between two lines

Method 36.1 — angle between two lines

The angle between two lines is found from the angle between their direction vectors $\mathbf{d}_1,\mathbf{d}_2$, using $\cos\theta=\dfrac{\mathbf{d}_1\cdot\mathbf{d}_2}{|\mathbf{d}_1||\mathbf{d}_2|}$. Since a line has no inherent "direction" (it extends both ways), the angle between two lines is always taken as the acute angle, so use $|\cos\theta|$ if the raw calculation gives an obtuse value.

Two straight lines crossing at a point P, with their direction vectors d1 and d2 drawn from P and the acute angle theta between them marked with an arc

Fig. 1 — the angle $\theta$ between two lines $l_1$ and $l_2$ is the acute angle between their direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$ at the point $P$ where they cross.

Example 1 · Angle between two lines

Find the acute angle between the lines with direction vectors $(2,1,-2)$ and $(1,-2,2)$.

Dot product: $2(1)+1(-2)+(-2)(2)=2-2-4=-4$.

$|\mathbf{d}_1|=3$, $|\mathbf{d}_2|=3$.

$\cos\theta=\dfrac{-4}{9}$; taking the acute angle, $\cos\theta=\dfrac{4}{9} \Rightarrow \theta=63.6°$ (3 s.f.).

Angle between a line and a plane

Theorem 36.2 — angle between a line and a plane

Let $\phi$ be the angle between a line (direction $\mathbf{d}$) and a plane (normal $\mathbf{n}$). Since the normal is perpendicular to the plane, the angle between $\mathbf{d}$ and $\mathbf{n}$ is the complement of $\phi$: $$\sin\phi=\frac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$$ (Using $\sin$, not $\cos$, and using the modulus to ensure $\phi$ is taken between $0°$ and $90°$.)

ProofWhy sine, not cosine, is used here

Let $\alpha$ be the angle between $\mathbf{d}$ and $\mathbf{n}$, so $\cos\alpha=\dfrac{\mathbf{d}\cdot\mathbf{n}}{|\mathbf{d}||\mathbf{n}|}$.

Standard scalar-product angle formula.

The normal is perpendicular to the plane, so $\alpha$ and $\phi$ (the line-plane angle) satisfy $\alpha+\phi=90°$.

Geometric relationship: the normal is $90°$ from anything lying within the plane, and the line makes angle $\phi$ with the plane itself.

$\cos\alpha=\cos(90°-\phi)=\sin\phi$

Standard complementary angle identity.

$\sin\phi=\dfrac{|\mathbf{d}\cdot\mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$

Modulus ensures $\phi$, which is always taken between $0°$ and $90°$, comes out non-negative.

∎

A line piercing a shaded plane at point F, with the normal vector n drawn perpendicular to the plane, a right angle marked between the normal and the surface, and the angle phi between the line and the plane marked with an arc

Fig. 2 — the line meets the plane at $F$. Because the normal $\mathbf{n}$ is perpendicular to the plane, the line-plane angle $\phi$ is the complement of the angle between $\mathbf{d}$ and $\mathbf{n}$ — which is exactly why $\sin$, not $\cos$, appears in the formula.

Example 2 · Angle between a line and a plane

Find the angle between the line $\mathbf{r}=(1,0,2)+\lambda(2,1,-2)$ and the plane $x-2y+2z=5$.

Direction $\mathbf{d}=(2,1,-2)$; normal $\mathbf{n}=(1,-2,2)$.

$\mathbf{d}\cdot\mathbf{n}=2(1)+1(-2)+(-2)(2)=2-2-4=-4$.

$|\mathbf{d}|=3$, $|\mathbf{n}|=3$.

$\sin\phi=\dfrac{|-4|}{9}=\dfrac49 \Rightarrow \phi=26.4°$ (3 s.f.).

Angle between two planes

Theorem 36.3 — angle between two planes

The angle between two planes equals the angle between their normal vectors $\mathbf{n}_1,\mathbf{n}_2$: $$\cos\theta=\frac{|\mathbf{n}_1\cdot\mathbf{n}_2|}{|\mathbf{n}_1||\mathbf{n}_2|}$$ (Using cosine here, like the line-line case, since we're comparing two normals directly — and the modulus again ensures the acute angle.)

Example 3 · Angle between two planes

Find the angle between the planes $2x+y-2z=5$ and $x-2y+2z=3$.

$\mathbf{n}_1=(2,1,-2)$, $\mathbf{n}_2=(1,-2,2)$.

$\mathbf{n}_1\cdot\mathbf{n}_2=2(1)+1(-2)+(-2)(2)=2-2-4=-4$.

$|\mathbf{n}_1|=3$, $|\mathbf{n}_2|=3$.

$\cos\theta=\dfrac{|-4|}{9}=\dfrac49 \Rightarrow \theta=63.6°$ (3 s.f.).

Example 4 · Recognising a line parallel to a plane

Show that the line $\mathbf{r}=(1,1,0)+\lambda(3,1,1)$ is parallel to the plane $x-2y-z=4$.

Direction $\mathbf{d}=(3,1,1)$; normal $\mathbf{n}=(1,-2,-1)$.

$\mathbf{d}\cdot\mathbf{n}=3(1)+1(-2)+1(-1)=3-2-1=0$.

Since $\mathbf{d}\cdot\mathbf{n}=0$, the direction of the line is perpendicular to the normal, meaning the line runs parallel to the plane (or lies within it).

Common mistake

Using $\cos$ instead of $\sin$ for a line-plane angle (or vice versa) is the most common error in this topic — remember: line vs line, and plane vs plane, both use cosine directly on the given vectors (direction or normal); line vs plane uses sine, because a direction vector and a normal vector are being compared, not two vectors that lie in the same kind of geometric object.

Exam-style questions

Question 1 2 marks

Find the angle between the line $\mathbf{r}=(0,1,2)+\lambda(1,2,-2)$ and the plane $2x-y+2z=7$.

$\mathbf{d}=(1,2,-2)$, $\mathbf{n}=(2,-1,2)$ (B1)

$\mathbf{d}\cdot\mathbf{n}=2-2-4=-4$; $|\mathbf{d}|=3$, $|\mathbf{n}|=3$ (M1, A1)

$\sin\phi=\dfrac{4}{9} \Rightarrow \phi=26.4°$ (3 s.f.) (M1, A1, A1)

Question 2 2 marks

Find the angle between the planes $x+2y-z=6$ and $3x-y+z=2$.

$\mathbf{n}_1=(1,2,-1)$, $\mathbf{n}_2=(3,-1,1)$ (B1)

$\mathbf{n}_1\cdot\mathbf{n}_2=3-2-1=0$ (M1, A1)

Since the dot product is $0$, the planes are perpendicular: $\theta=90°$. (A1, B1, B1)

Question 3 2 marks

Find the acute angle between the lines with direction vectors $(1,1,1)$ and $(2,-1,1)$, to 1 decimal place.

Dot product: $2-1+1=2$ (M1, A1)

$|\mathbf{d}_1|=\sqrt3$, $|\mathbf{d}_2|=\sqrt6$ (A1)

$\cos\theta=\dfrac{2}{\sqrt{18}}=\dfrac{2}{3\sqrt2}=0.4714\ldots \Rightarrow \theta=61.9°$ (3 s.f.) (M1, A1, A1)

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Points of Intersection