Vectors, Worked Examples (KBAT)

Reach $P$ by going from $O$ to $A$, then a fraction of the way along $AB$: $\overrightarrow{OP}=\overrightarrow{OA}+\overrightarrow{AP}$. First write $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$:

$$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf{b}-\mathbf{a}$$

Since $AP:PB=1:2$, the point $P$ is one part along a line split into three, so $\overrightarrow{AP}=\tfrac{1}{3}\overrightarrow{AB}$:

$$\overrightarrow{AP}=\tfrac{1}{3}(\mathbf{b}-\mathbf{a})$$

Now add along the route and collect like terms:

$$\overrightarrow{OP}=\mathbf{a}+\tfrac{1}{3}(\mathbf{b}-\mathbf{a})=\mathbf{a}-\tfrac{1}{3}\mathbf{a}+\tfrac{1}{3}\mathbf{b}=\tfrac{2}{3}\mathbf{a}+\tfrac{1}{3}\mathbf{b}$$

First locate $M$ and $N$ as vectors from $O$: $\overrightarrow{OM}=\tfrac{1}{2}\mathbf{a}$, and since $ON:NB=2:1$, $\overrightarrow{ON}=\tfrac{2}{3}\mathbf{b}$.

Follow the route $O\to A\to P$. Using $\overrightarrow{AN}=\overrightarrow{ON}-\overrightarrow{OA}=\tfrac{2}{3}\mathbf{b}-\mathbf{a}$:

$$\overrightarrow{OP}=\mathbf{a}+h\!\left(\tfrac{2}{3}\mathbf{b}-\mathbf{a}\right)=(1-h)\mathbf{a}+\tfrac{2h}{3}\mathbf{b}$$

Now follow the route $O\to B\to P$. Using $\overrightarrow{BM}=\overrightarrow{OM}-\overrightarrow{OB}=\tfrac{1}{2}\mathbf{a}-\mathbf{b}$:

$$\overrightarrow{OP}=\mathbf{b}+k\!\left(\tfrac{1}{2}\mathbf{a}-\mathbf{b}\right)=\tfrac{k}{2}\mathbf{a}+(1-k)\mathbf{b}$$

Because $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the two expressions for $\overrightarrow{OP}$ can only match if their coefficients match. Equate them:

$$1-h=\tfrac{k}{2}\qquad\text{and}\qquad \tfrac{2h}{3}=1-k$$

From the first equation $k=2(1-h)=2-2h$. Substitute into the second:

$$\tfrac{2h}{3}=1-(2-2h)=2h-1 \;\Rightarrow\; 2h=6h-3 \;\Rightarrow\; 4h=3 \;\Rightarrow\; h=\tfrac{3}{4}$$

$$k=2-2\!\left(\tfrac{3}{4}\right)=2-\tfrac{3}{2}=\tfrac{1}{2}$$

First write the vector $\mathbf{a}+t\mathbf{b}$ in components, keeping $t$ as an unknown:

$$\mathbf{a}+t\mathbf{b}=\binom{3+t}{-4+2t}$$

The magnitude condition $|\mathbf{a}+t\mathbf{b}|=2\sqrt{10}$ is easier to use once squared, since $(2\sqrt{10})^{2}=40$. Set the sum of squared components equal to $40$:

$$(3+t)^{2}+(-4+2t)^{2}=40$$

Expand each square carefully and collect terms:

$$(9+6t+t^{2})+(16-16t+4t^{2})=5t^{2}-10t+25=40$$

Bring everything to one side and divide by $5$ to get a simple quadratic, then factorise:

$$5t^{2}-10t-15=0 \;\Rightarrow\; t^{2}-2t-3=0 \;\Rightarrow\; (t-3)(t+1)=0$$

Since $OC:CA=2:1$, point $C$ is $\tfrac{2}{3}$ of the way from $O$ to $A$; the same fraction places $D$ on $OB$. Write both as vectors from $O$:

$$\overrightarrow{OC}=\tfrac{2}{3}\mathbf{a}\qquad \overrightarrow{OD}=\tfrac{2}{3}\mathbf{b}$$

Find $\overrightarrow{CD}$ by subtracting, and compare it with $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$:

$$\overrightarrow{CD}=\overrightarrow{OD}-\overrightarrow{OC}=\tfrac{2}{3}\mathbf{b}-\tfrac{2}{3}\mathbf{a}=\tfrac{2}{3}(\mathbf{b}-\mathbf{a})=\tfrac{2}{3}\overrightarrow{AB}$$

Travelling $A\to B\to C\to A$ returns to the start, so the three displacement vectors sum to the zero vector. Use this to isolate $\overrightarrow{CA}$:

$$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\mathbf{0}\;\Rightarrow\;\overrightarrow{CA}=-\left(\overrightarrow{AB}+\overrightarrow{BC}\right)=-\left[\binom{5}{-3}+\binom{-2}{7}\right]=-\binom{3}{4}=\binom{-3}{-4}$$

Now find the magnitude of $\overrightarrow{CA}$ and divide by it to get a unit vector:

$$|\overrightarrow{CA}|=\sqrt{(-3)^{2}+(-4)^{2}}=\sqrt{25}=5\;\Rightarrow\;\hat{\mathbf{CA}}=\dfrac{1}{5}\binom{-3}{-4}$$

In parallelogram $ABCD$, the opposite sides $AB$ and $DC$ are equal and parallel, so $\overrightarrow{AB}=\overrightarrow{DC}$. First find $\overrightarrow{AB}$:

$$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\binom{5}{3}-\binom{1}{2}=\binom{4}{1}$$

Since $\overrightarrow{DC}=\overrightarrow{OC}-\overrightarrow{OD}$ must equal this, rearrange for $\overrightarrow{OD}$:

$$\overrightarrow{OD}=\overrightarrow{OC}-\overrightarrow{AB}=\binom{6}{7}-\binom{4}{1}=\binom{2}{6}$$

Because $\overrightarrow{OA}$ is parallel to $3\mathbf{i}+4\mathbf{j}$, it must be a scalar multiple of it: $\overrightarrow{OA}=k(3\mathbf{i}+4\mathbf{j})$ for some scalar $k$, so $p=3k$ and $q=4k$.

$$\overrightarrow{OA}=k(3\mathbf{i}+4\mathbf{j})=3k\mathbf{i}+4k\mathbf{j}$$

Use the magnitude condition, noting $|3\mathbf{i}+4\mathbf{j}|=\sqrt{3^{2}+4^{2}}=5$, to find $k$:

$$|\overrightarrow{OA}|=|k|\sqrt{3^{2}+4^{2}}=5|k|=20\;\Rightarrow\;|k|=4\;\Rightarrow\;k=4\text{ or }k=-4$$

Three points are collinear when the vector between two of them is a scalar multiple of the vector between another pair sharing a common point. Find $\overrightarrow{LM}$ and $\overrightarrow{MN}$:

$$\overrightarrow{LM}=\overrightarrow{OM}-\overrightarrow{OL}=\binom{5}{4}-\binom{2}{1}=\binom{3}{3}\qquad \overrightarrow{MN}=\overrightarrow{ON}-\overrightarrow{OM}=\binom{11}{10}-\binom{5}{4}=\binom{6}{6}$$

Compare the two vectors:

$$\overrightarrow{MN}=\binom{6}{6}=2\binom{3}{3}=2\overrightarrow{LM}$$