Review question

# Can we find the angle between the planes $ABC$ and $ABD$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6449

## Solution

The position vectors of four points $A$, $B$, $C$, $D$ relative to an origin $O$ are given below. The vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are mutually perpendicular unit vectors. \begin{align*} A \colon\;& 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}, \\ B \colon\;& \phantom{2}\mathbf{i} + \phantom{3}\mathbf{j} - \mathbf{k}, \\ C \colon\;& \phantom{2}\mathbf{i} \phantom{{}+{}3\mathbf{i}} + \mathbf{k}, \\ D \colon\;& \phantom{2\mathbf{i}{}+{}} 3\mathbf{j}.\phantom{{}+{}\mathbf{k}} \end{align*}

Find

1. the equation (in any form) of the line $AB$,

We can write the position vectors $A$ to $D$ in column form as $\mathbf{a} = \begin{pmatrix}2 \\3\\1\end{pmatrix}$, $\mathbf{b} = \begin{pmatrix}1 \\1\\-1\end{pmatrix}$, $\mathbf{c} = \begin{pmatrix}1 \\0\\1\end{pmatrix}$ and $\mathbf{d} = \begin{pmatrix}0 \\3\\0\end{pmatrix}$.

Writing vectors in column form rather than $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ form for the calculations we are going to be doing reduces the likelihood of us mixing up the components of our vectors when we combine them.

A direction vector of the line $AB$ is $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}1 \\1\\-1\end{pmatrix}-\begin{pmatrix}2 \\3\\1\end{pmatrix} = \begin{pmatrix}-1\\-2\\-2\end{pmatrix}.$

To avoid so many minus signs, we will use the direction vector $\mathbf{p}=-\overrightarrow{AB}=\begin{pmatrix}1\\2\\2\end{pmatrix}$ instead.

Any point on the line $AB$ has position vector $\mathbf{r}=\mathbf{a} + s\mathbf{p}$ for some real value of $s$. Hence an equation of the line $AB$ is $\mathbf{r}= \begin{pmatrix}2 \\3\\1\end{pmatrix} + s\begin{pmatrix}1\\2\\2\end{pmatrix}$, where $s$ is any real number.

1. the shortest distance between $AB$ and $CD$,

Using the same method as above, we let $\mathbf{q}=-\overrightarrow{CD}=-(\mathbf{d}-\mathbf{c})=\begin{pmatrix}1\\-3\\1\end{pmatrix}$ (again, aiming to have fewer minus signs). Then an equation of the line $CD$ is $\mathbf{r}= \mathbf{c}+t\mathbf{q}=\begin{pmatrix}1\\0\\1\end{pmatrix} + t\begin{pmatrix}1\\-3\\1\end{pmatrix}$, where $t$ is any real number.

Why have we used $t$ here and not $s$ again? On the other hand, we have used $\mathbf{r}$ again; is this a problem?

Let’s suppose $E$ is on $AB$ and $F$ is on $CD$ with $EF$ having the shortest distance between $AB$ and $CD$. To help us understand this part, we’ll draw a quick sketch of the situation.

(Note that we have sketched $E$ as lying between $A$ and $B$, but this may or may not be the case, and similarly with $F$. It makes no difference to the arguments we will use.)

Because $EF$ has the shortest distance between the two lines, it must be perpendicular to both of them. (Otherwise, if we fix $E$, we can make $EF$ shorter by moving $F$ until $EF$ is perpendicular to $CD$, and likewise with moving $E$.)

There are at least two very different ways to proceed at this point. We could either use scalar (dot) products or vector (cross) products. We will use scalar products here, and then we will show how the rest of this question could be answered in an alternative way using vector products.

Now we can write the position vectors of $E$ and $F$ as $\mathbf{e}= \begin{pmatrix}2+s \\3+2s\\1+2s\end{pmatrix}$, $\mathbf{f}= \begin{pmatrix}1+t\\-3t\\1+t\end{pmatrix}$ for some values of $s$ and $t$. This gives $\overrightarrow{EF} = \mathbf{f}-\mathbf{e} = \begin{pmatrix}1+t\\-3t\\1+t\end{pmatrix}-\begin{pmatrix}2+s \\3+2s\\1+2s\end{pmatrix}= \begin{pmatrix}-1+t-s \\-3t-3-2s\\t-2s\end{pmatrix}.$

Since $\overrightarrow{EF}$ is perpendicular to both $AB$ and $CD$, we can obtain two equations by noting that the scalar product of $\overrightarrow{EF}$ with both direction vectors $\mathbf{p}$ and $\mathbf{q}$ is zero: \begin{align*} \overrightarrow{EF}.\mathbf{p}&=\begin{pmatrix}-1+t-s \\-3t-3-2s\\t-2s\end{pmatrix}. \begin{pmatrix}1\\2\\2\end{pmatrix}=(-1+t-s)+2(-3t-3-2s)+2(t-2s)=0\\ \overrightarrow{EF}.\mathbf{q}&=\begin{pmatrix}-1+t-s \\-3t-3-2s\\t-2s\end{pmatrix}. \begin{pmatrix}1\\-3\\1\end{pmatrix}=(-1+t-s)-3(-3t-3-2s)+(t-2s)=0 \end{align*}

Simplifying these gives $7+9s+3t=0$ and $8+3s+11t=0$. Solving these simultaneous equations in $s$ and $t$, we get $s=-\dfrac{53}{90}$, $t=-\dfrac{17}{30}$, so that $\cmarraystretch{2.2}\overrightarrow{EF} = \begin{pmatrix}-1+\dfrac{-17}{30}+\dfrac{53}{90}\\-3\dfrac{-17}{30}-3+2\dfrac{53}{90}\\\dfrac{-17}{30}+2\dfrac{53}{90}\end{pmatrix} =\begin{pmatrix}-\dfrac{44}{45} \\-\dfrac{11}{90}\\\dfrac{11}{18}\end{pmatrix}.$

The magnitude of $\overrightarrow{EF}$ is therefore $\sqrt{\left(-\dfrac{44}{45}\right)^2+\left(-\dfrac{11}{90}\right)^2+\left(\dfrac{11}{18}\right)^2} = \dfrac{11}{90}\sqrt{8^2+1^2+5^2} =\dfrac{11\sqrt{10}}{30}$, and this is the distance between the lines $AB$ and $CD$.

1. the equation (in any form) of the plane $ABC$,

We’ll aim for the Cartesian equation of the plane $ABC$ in the form $px+qy+rx+u =0$.

We do this with an eye on the last part of the question. The angle between two planes is the angle between the normals to each plane, and the normal to a plane is easily found from its Cartesian equation.

The point $\mathbf{a} = \begin{pmatrix}2\\3\\1\end{pmatrix}$ is on the plane, so $2p+3q+r +u= 0\quad (1)$.

The point $\mathbf{b} = \begin{pmatrix}1 \\1\\-1\end{pmatrix}$ is on the plane, so $p+q-r +u= 0\quad (2)$.

The point $\mathbf{c} = \begin{pmatrix}1 \\0\\1\end{pmatrix}$ is on the plane, so $p+r +u= 0\quad (3)$.

We can eliminate the $u$ terms by subtracting pairs of equations:

\begin{align*} (1)-(2) \implies& p+2q+2r =0\\ (2)-(3) \implies& \phantom{p+2}q-2r =0 \end{align*}

Adding these then gives $p+3q=0$, so $p=-3q$, $r=\frac12 q$ and hence $u=\frac52 q$.

Thus the equation of the plane $ABC$ is $-3qx+qy+\frac12 qz+\frac52 q=0$, or $-6x+2y+z+5=0$ (dividing by $\frac12 q$; we can multiply or divide a Cartesian equation of the plane by any non-zero constant we wish without changing the plane represented).

1. the angle between the planes $ABC$ and $ABD$.

Using a similar method to that above, we can find the equation of the plane $ABD$.

Once again, we’ll aim for the Cartesian equation of the plane in the form $px+qy+rx+u =0$.

The point $\mathbf{a} = \begin{pmatrix}2\\3\\1\end{pmatrix}$ is on the plane, so $2p+3q+r +u= 0\quad (1)$.

The point $\mathbf{b} = \begin{pmatrix}1 \\1\\-1\end{pmatrix}$ is on the plane, so $p+q-r +u= 0\quad (2)$.

The point $\mathbf{d} = \begin{pmatrix}0 \\3\\0\end{pmatrix}$ is on the plane, so $3q +u= 0\quad (3)$.

Again, we can eliminate $u$ by subtracting pairs of equations (we can always do this with this type of system of equations):

\begin{align*} (1)-(2) \implies& \phantom{2}p+2q+2r =0\\ (1)-(3) \implies& 2p\phantom{{}+2q}+\phantom{2}r =0\\ \end{align*}

Thus $r=-2p$, $q=\frac32 p$ and hence $u=-\frac92 p$.

Thus the equation of the plane $ABD$ is $px+\frac32 py-2pz-\frac92 p=0$ or $2x+3y-4z-9=0$.

Hence a normal to the plane $ABC$ is $\begin{pmatrix}-6 \\2\\1\end{pmatrix}$, while a normal to the plane $ABD$ is $\begin{pmatrix}2 \\3\\-4\end{pmatrix}$.

To find the angle between these two vectors, we use the scalar product. We have $\begin{pmatrix}-6 \\2\\1\end{pmatrix}\cdot \begin{pmatrix}2 \\3\\-4\end{pmatrix} = -12+6-4=-10$.

We also have $\left \vert \begin{pmatrix}-6 \\2\\1\end{pmatrix} \right \vert = \sqrt{36+4+1}$, while $\left \vert \begin{pmatrix}2 \\3\\-4\end{pmatrix} \right \vert = \sqrt{4+9+16}$.

Thus $\cos \theta = \dfrac{-10}{\sqrt{41}\sqrt{29}} = -0.2900\ldots$, so the angle between the normals is $\theta = 106.9^\circ$ (to 1 d.p.). The angle between the planes is equal to the angle between the normals, so the (obtuse) angle between the planes is $106.9^\circ$. If we wish to give the acute angle between the planes instead, this is then $180^\circ-106.9^\circ=73.1^\circ$ (to 1 d.p.).