Angles between planes and between lines and planesAQA A-Level Further Maths: Revision notes

A plane with Cartesian equation $ax+by+cz=d$ has normal vector $\mathbf n=\begin{pmatrix}a \\ b \\ c\end{pmatrix}$, which is perpendicular to every direction lying in the plane. A line $\mathbf r=\mathbf a+t\mathbf b$ has direction vector $\mathbf b$. Every angle in this topic is found by applying the scalar product $\mathbf u\cdot\mathbf v=|\mathbf u||\mathbf v|\cos\theta$ to normals and direction vectors, never to the plane itself. Rearranged, $\cos\theta=\dfrac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|}$.

The angle between two planes equals the angle between their normals. For planes with normals $\mathbf n_1$ and $\mathbf n_2$, $\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}.$ The modulus gives the acute angle, which is the one normally asked for. Example: $2x+y-2z=5$ and $2x+3y+6z=11$ have $\mathbf n_1\cdot\mathbf n_2=4+3-12=-5$, $|\mathbf n_1|=3$, $|\mathbf n_2|=7$, so $\cos\theta=\frac{5}{21}$ and $\theta=76.2^\circ$. If $\mathbf n_1\cdot\mathbf n_2=0$ the planes are perpendicular; if the normals are parallel the planes are parallel.

The angle $\theta$ between a line and a plane is measured from the line to the plane, so it is the complement of the angle between the line and the normal. Using direction vector $\mathbf b$ and normal $\mathbf n$, $\sin\theta=\frac{|\mathbf b\cdot\mathbf n|}{|\mathbf b||\mathbf n|}.$ Note sine, not cosine. Example: $\mathbf b=(2,-1,2)$ and $\mathbf n=(1,2,2)$ give $\mathbf b\cdot\mathbf n=4$, $|\mathbf b||\mathbf n|=9$, so $\sin\theta=\frac49$ and $\theta=26.4^\circ$ (the angle to the normal would be $63.6^\circ$). If $\mathbf b\cdot\mathbf n=0$ the line is parallel to the plane (or lies in it), and $\theta=0^\circ$.

A cuboid has edges along the axes with $A=(4,0,0)$, $C=(0,3,0)$, $D=(0,0,2)$. Plane $ACD$ is $3x+4y+6z=12$ (each point satisfies it, and three non-collinear points fix a plane). The base is $z=0$ with normal $(0,0,1)$.

• Plane $ACD$ and the base: $\cos\theta=\frac{6}{\sqrt{61}}$, so $\theta=39.8^\circ$.
• The line from $O$ to $G(4,3,2)$ has $\mathbf b=(4,3,2)$. With plane $ACD$: $\sin\theta=\frac{36}{\sqrt{29}\sqrt{61}}$, so $\theta=58.9^\circ$. With the base: $\sin\phi=\frac{2}{\sqrt{29}}$, so $\phi=21.8^\circ$.

• Plane and plane: normals, $\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}$.
• Line and plane: direction and normal, $\sin\theta=\frac{|\mathbf b\cdot\mathbf n|}{|\mathbf b||\mathbf n|}$.
• Line and line: two directions, $\cos\theta=\frac{|\mathbf b_1\cdot\mathbf b_2|}{|\mathbf b_1||\mathbf b_2|}$.

Give angles in degrees to 1 d.p. unless the question specifies radians, and keep exact values such as $\sqrt{61}$ until the last step.