Review question

# What is the position vector of $D$ if $ABCD$ is a parallelogram? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8215

## Solution

The position vectors of points $A$, $B$ and $C$ are $\begin{equation*} \mathbf{a} = 4\mathbf{i} - 9\mathbf{j} - \mathbf{k}, \quad \mathbf{b} = \mathbf{i} + 3\mathbf{j} + 5\mathbf{k}, \quad\text{and}\quad \mathbf{c} = p\mathbf{i} - \mathbf{j} + 3\mathbf{k}. \end{equation*}$
1. Find the unit vector parallel to the vector $\mathbf{AB}$.

The vector $\overrightarrow{AB}$ begins at $A$ and ends at $B$. We have the following diagram.

From the diagram, we have that $\mathbf{a} + \overrightarrow{AB} = \mathbf{b}$, i.e. that $\begin{equation*} \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}) - (4\mathbf{i} - 9\mathbf{j} - \mathbf{k}) = -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}. \end{equation*}$ The length of this vector is $\begin{equation*} \sqrt{(-3)^2 + 12^2 + 6^2} = \sqrt{9 + 144 + 36} = \sqrt{189} = 3\sqrt{21}, \end{equation*}$ and therefore the unit vector parallel to $\overrightarrow{AB}$ is \begin{align*} &\frac{-3}{3\sqrt{21}} \mathbf{i} + \frac{12}{3\sqrt{21}} \mathbf{j} + \frac{6}{3\sqrt{21}} \mathbf{k} \\ =&\frac{-1}{\sqrt{21}} \mathbf{i} + \frac{4}{\sqrt{21}} \mathbf{j} + \frac{2}{\sqrt{21}} \mathbf{k}. \end{align*}
1. Find the value of $p$ such that $A$, $B$ and $C$ are collinear.

The points $A, B$ and $C$ are collinear $\iff \overrightarrow{AB}$ is parallel to $\overrightarrow{AC} \iff \overrightarrow{AB} = k\overrightarrow{AC}$ for some constant $k$.

Now

$\overrightarrow{AB} = -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k},\quad \overrightarrow{AC} = (p-4) \mathbf{i} + 8 \mathbf{j} + 4 \mathbf{k},$

and thus

$2\overrightarrow{AB} = -6 \mathbf{i} + 24 \mathbf{j} + 12 \mathbf{k},\quad 3\overrightarrow{AC} = 3(p-4) \mathbf{i} + 24 \mathbf{j} + 12 \mathbf{k}.$

Comparing coefficients, we can say that $A, B$ and $C$ are collinear if and only if $-6 = 3(p-4)$, which is true if and only if $p = 2$.

Alternatively, we could you use the fact that $C$ lies on the straight line $AB$. We could find the equation of the line $AB$ and then substitute $C$ into this equation to find $p$.

1. If $p = -2$, find the position vector of $D$ so that $ABCD$ is a parallelogram.

We have the following sketch.

The position vector of $D$ is therefore \begin{align*} \mathbf{d} = \mathbf{a} + \overrightarrow{AD} = \mathbf{a} + \overrightarrow{BC}&= \mathbf{a} + \mathbf{c} - \mathbf{b} \\ &= \left( 4\mathbf{i} - 9\mathbf{j} - \mathbf{k} \right) + \left( -2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \right) - \left( \mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \right) \\ &= (4 - 2 - 1) \mathbf{i} + (-9 - 1 - 3) \mathbf{j} + (-1 + 3 - 5) \mathbf{k} \\ &= \mathbf{i} - 13 \mathbf{j} - 3 \mathbf{k}. \end{align*}
Check: does $\overrightarrow{DC} = \overrightarrow{AB}$? We have that \begin{align*} \overrightarrow{AB} &= \mathbf{b} - \mathbf{a} \\ &= \left( \mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \right) - \left( 4\mathbf{i} - 9\mathbf{j} - \mathbf{k} \right) \\ &= -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}, \end{align*} and that \begin{align*} \overrightarrow{DC} &= \mathbf{c} - \mathbf{d} \\ &= \left( -2\mathbf{i} - \mathbf{j} + 3\mathbf{k} \right) - \left( \mathbf{i} - 13 \mathbf{j} - 3 \mathbf{k} \right) \\ &= -3 \mathbf{i} + 12 \mathbf{j} + 6 \mathbf{k}. \end{align*}
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