Combining Vectors

Can we find the angle between the planes $ABC$ and $ABD$?
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Ref: R6449

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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

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.

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.

the lines a b and c d with the points e and f shown and joined indicating that e f meets a b and c d at right angles — Figure 1

(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 $CD$.

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).

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.).

If we draw a sketch of the situation, we may see another possible way of thinking about the angle between these two planes. Though this is less straightforward than the above approach, especially after having done part (iii), it is perhaps more informative. It is also interesting in that it does not require us to work out the equations of the planes.

The planes A B C and A B D shown with the vectors B C and B D shown with their components as described in the text

The two planes $ABC$ and $ABD$ meet along the line $AB$. We have also drawn the unit vector $\hat{\mathbf{p}}$ (in the direction of $\overrightarrow{BA}$), which is $\mathbf{p}/|\mathbf{p}|=\frac{1}{3}\mathbf{p}=\frac{1}{3}\begin{pmatrix}1\\2\\2\end{pmatrix}$.

The angle between the planes is the angle between any two vectors $\overrightarrow{BC'}$ and $\overrightarrow{BD'}$ which lie in the planes $ABC$ and $ABD$ respectively, and which are both perpendicular to $AB$. As shown in the sketch, we can choose $C'$ and $D'$ so that $\overrightarrow{BC'}$ is the component of $\overrightarrow{BC}$ perpendicular to $\overrightarrow{BA}$, and likewise for $\overrightarrow{BD'}$. We can work out $\overrightarrow{C'C}$: it is the component of $\overrightarrow{BC}$ parallel to $\hat{\mathbf{p}}$, so \[|C'C|=\overrightarrow{BC\vphantom{C'}}\cdot\hat{\mathbf{p}}=\begin{pmatrix}0\\-1\\2\end{pmatrix}\cdot\frac{1}{3} \begin{pmatrix}1\\2\\2\end{pmatrix} = \frac{2}{3}\] giving \[\overrightarrow{C'C}=\frac{2}{3}\hat{\mathbf{p}} = \frac{2}{9}\begin{pmatrix}1\\2\\2\end{pmatrix}.\] Thus \[\overrightarrow{BC'}=\overrightarrow{BC\vphantom{C'}}+\overrightarrow{CC'}=\begin{pmatrix}0\\-1\\2\end{pmatrix}-\frac{2}{9}\begin{pmatrix}1\\2\\2\end{pmatrix} =\frac{1}{9}\begin{pmatrix}-2\\-13\\14\end{pmatrix}.\] Similarly, \[\overrightarrow{D'D}=(\overrightarrow{BD\vphantom{D'}}\cdot\hat{\mathbf{p}})\;\hat{\mathbf{p}}=\left(\begin{pmatrix}-1\\2\\1\end{pmatrix}\cdot\frac{1}{3} \begin{pmatrix}1\\2\\2\end{pmatrix}\right)\;\hat{\mathbf{p}} = \frac{5}{9}\begin{pmatrix}1\\2\\2\end{pmatrix},\] giving \[\overrightarrow{BD'}=\overrightarrow{BD\vphantom{D'}}+\overrightarrow{DD'}=\begin{pmatrix}-1\\2\\1\end{pmatrix}-\frac{5}{9}\begin{pmatrix}1\\2\\2\end{pmatrix} =\frac{1}{9}\begin{pmatrix}-14\\8\\-1\end{pmatrix}.\]

Then the angle $\theta$ between $\overrightarrow{BC'}$ and $\overrightarrow{BD'}$ is found using the scalar product. For simplicity, we may as well use $9\overrightarrow{BC'}$ and $9\overrightarrow{BD'}$, as this does not change the angle. We have \[9\overrightarrow{BC'}\cdot 9\overrightarrow{BD'} = -90, \quad |9\overrightarrow{BC'}|=3\sqrt{41}\quad\text{and}\quad |9\overrightarrow{BD'}|=3\sqrt{29}\] and hence $\cos\theta=-\dfrac{90}{3\sqrt{41}.3\sqrt{29}}=-\dfrac{10}{\sqrt{41}\sqrt{29}}$ as before.

We can answer parts (ii)–(iv) in a different way using the vector product.

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

For part (ii), starting from Figure 1 in (ii) above, we see that the vector $\overrightarrow{EF}$ is perpendicular to both $\mathbf{p}$ and $\mathbf{q}$, so it must be parallel to $\mathbf{p}\times\mathbf{q}$. We can therefore redraw this figure as follows:

the lines a b and c d with the points e and f shown and joined with the vector k p cross q indicating that e f meets a b and c d at right angles

Thus to find the length of $EF$, we can take the scalar product of $\overrightarrow{EF}$ with $\mathbf{p}\times\mathbf{q}$: \[\overrightarrow{EF}\cdot(\mathbf{p}\times\mathbf{q})=k(\mathbf{p}\times\mathbf{q})\cdot(\mathbf{p}\times\mathbf{q})=k|\mathbf{p}\times\mathbf{q}|^2\] so that \[|\overrightarrow{EF}|=|k(\mathbf{p}\times\mathbf{q})|=\frac{|\overrightarrow{EF}\cdot(\mathbf{p}\times\mathbf{q})|}{|\mathbf{p}\times\mathbf{q}|}\] (where we need the absolute value signs in the numerator because $k$ might be negative).

We can now work out $\overrightarrow{EF}\cdot(\mathbf{p}\times\mathbf{q})$ itself: \[\begin{align*} \overrightarrow{EF}\cdot(\mathbf{p}\times\mathbf{q})&=(\mathbf{f}-\mathbf{e})\cdot(\mathbf{p}\times\mathbf{q})\\ &=(\mathbf{c}+t\mathbf{q}-\mathbf{a}-s\mathbf{p})\cdot(\mathbf{p}\times\mathbf{q})\\ &=(\mathbf{c}-\mathbf{a})\cdot(\mathbf{p}\times\mathbf{q}). \end{align*}\]

On the last line, we have used the result that $\mathbf{p}\cdot(\mathbf{p}\times\mathbf{q})=\mathbf{q}\cdot(\mathbf{p}\times\mathbf{q})=0$ since $\mathbf{p}\times\mathbf{q}$ is perpendicular to both $\mathbf{p}$ and $\mathbf{q}$.

Therefore the distance between the lines is \[EF=\frac{|(\mathbf{c}-\mathbf{a})\cdot(\mathbf{p}\times\mathbf{q})|}{|\mathbf{p}\times\mathbf{q}|}.\]

Note that this formula for the distance does not require us to work out the actual position vector of either $E$ or $F$!

Now \[\mathbf{p}\times\mathbf{q}=\begin{pmatrix}1\\2\\2\end{pmatrix}\times \begin{pmatrix}1\\-3\\1\end{pmatrix} = \begin{pmatrix} 8\\ 1\\ -5\end{pmatrix}\] so $|\mathbf{p}\times\mathbf{q}|=\sqrt{90}=3\sqrt{10}$, and hence \[EF=\frac{1}{3\sqrt{10}}\left|\left(\begin{pmatrix}1 \\0\\1\end{pmatrix} - \begin{pmatrix}2 \\3\\1\end{pmatrix}\right)\cdot\begin{pmatrix} 8\\ 1\\ -5\end{pmatrix}\right|=\frac{11}{3\sqrt{10}}.\]

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

The points $A$, $B$ and $C$ lie in this plane, so the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ lie within the plane. Hence a normal $\mathbf{n}$ to the plane can be found by computing $\overrightarrow{AB}\times\overrightarrow{AC}$:

\[\begin{align*} \mathbf{n} &= \overrightarrow{AB}\times\overrightarrow{AC}\\ &= (\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a})\\ &= \begin{pmatrix}-1 \\-2\\-2\end{pmatrix} \times \begin{pmatrix}-1 \\-3\\0\end{pmatrix}\\ &= \begin{pmatrix}-6 \\ 2\\ 1\end{pmatrix}. \end{align*}\]

The equation of the plane is then $\mathbf{r}\cdot\mathbf{n}=c$ for some constant $c$; by substituting in $\mathbf{r}=\mathbf{a}$, for example, we can deduce $c=\mathbf{a}\cdot\mathbf{n}=-5$, and hence the plane $ABC$ has equation $\mathbf{r}\cdot\mathbf{n}=-5$.

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

We can likewise find a normal $\mathbf{n}'$ to the plane $ABD$:

\[\begin{align*} \mathbf{n}' &= \overrightarrow{AB}\times\overrightarrow{AD}\\ &= (\mathbf{b}-\mathbf{a})\times(\mathbf{d}-\mathbf{a})\\ &= \begin{pmatrix}-1 \\-2\\-2\end{pmatrix} \times \begin{pmatrix}-2 \\0\\ -1\end{pmatrix}\\ &= \begin{pmatrix}2 \\ 3\\ -4\end{pmatrix}. \end{align*}\]

Then the angle between the planes equals the angle between the normals $\mathbf{n}$ and $\mathbf{n}'$, which we can work out as in the first approach.

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Combining Vectors

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

Solution

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Lines

Can we find the angle between the planes $ABC$ and $ABD$?
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