Planes and the scalar productEdexcel A-Level Further Maths: Revision notes

A plane is fixed by a point and two non-parallel directions lying in it: $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c,$ where $\mathbf a$ is the position vector of a point on the plane, $\mathbf b$ and $\mathbf c$ are non-parallel direction vectors in the plane, and $\lambda$, $\mu$ are parameters. Through three points $A$, $B$, $C$, use $\mathbf a=\overrightarrow{OA}$, $\mathbf b=\overrightarrow{AB}$ and $\mathbf c=\overrightarrow{AC}$. Changing the point or the pair of directions gives other correct equations of the same plane.

For $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$: $\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta,$ where $\theta$ is the angle between the vectors. The result is a number, not a vector. So $\cos\theta=\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}$. Example: $\mathbf p=(3,2,-1)$, $\mathbf q=(1,-4,2)$: $\mathbf p\cdot\mathbf q=3-8-2=-7$, $|\mathbf p||\mathbf q|=\sqrt{14}\sqrt{21}$, so $\cos\theta=-0.408$ and $\theta=114.1^\circ$. Perpendicular vectors (non-zero) have $\mathbf a\cdot\mathbf b=0$, because $\cos90^\circ=0$.

A normal vector $\mathbf n$ is perpendicular to every direction in the plane. For a point $\mathbf a$ on the plane and any point $\mathbf r$ on it, $(\mathbf r-\mathbf a)\cdot\mathbf n=0$, so $\mathbf r\cdot\mathbf n=k,\qquad k=\mathbf a\cdot\mathbf n.$ Writing $\mathbf n=(a,b,c)$ and $\mathbf r=(x,y,z)$ gives the Cartesian equation $ax+by+cz=d$ with $d=k$; the coefficients of $x,y,z$ are the components of a normal. To find $\mathbf n$ from $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c$, solve $\mathbf n\cdot\mathbf b=0$ and $\mathbf n\cdot\mathbf c=0$ and choose any non-zero solution. Example: $\mathbf b=(1,1,0)$, $\mathbf c=(0,1,3)$ gives $n_1+n_2=0$ and $n_2+3n_3=0$, so $\mathbf n=(3,-3,1)$. With $\mathbf a=(1,0,2)$, $k=5$ and the plane is $3x-3y+z=5$.

To test whether a point lies on a plane, substitute its coordinates into the Cartesian equation. Any normal works: multiples such as $(6,-6,2)$ give the same plane with $6x-6y+2z=10$. From Cartesian to vector form: pick three non-collinear points on the plane (choose two coordinates and solve for the third), then use $\mathbf a$, $\overrightarrow{AB}$ and $\overrightarrow{AC}$. From vector to Cartesian, use the normal as above.

Two lines: use the direction vectors $\mathbf d_1$, $\mathbf d_2$ in $\cos\theta=\frac{|\mathbf d_1\cdot\mathbf d_2|}{|\mathbf d_1||\mathbf d_2|}$ for the acute angle. Two planes: the angle between the planes equals the acute angle between their normals, $\cos\theta=\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}$. Perpendicular planes have $\mathbf n_1\cdot\mathbf n_2=0$. Line and plane: find the acute angle $\alpha$ between the direction $\mathbf d$ and the normal $\mathbf n$, then the angle with the plane is $90^\circ-\alpha$. Equivalently $\sin\varphi=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}$. Example: $\mathbf d=(2,-1,2)$, $\mathbf n=(1,2,-2)$: $\mathbf d\cdot\mathbf n=-4$, so $\sin\varphi=\frac49$ and $\varphi=26.4^\circ$. To find where a line meets a plane, put the general point of the line into the plane's equation and solve for the parameter.