Solving geometric problems

In a nutshell

You can use vectors to solve geometric problems, including finding the position vector of a point that divides a line segment into a given ratio. You can prove the relationships between given shapes, lines or points, and find the angle in a given triangle. It helps to draw a diagram for yourself, and to figure out what information you need to solve the problem.

Finding the position vector

The point $$P$$ divides a line segment $$AB$$ into a ratio $$\lambda:\mu$$.

From the diagram, this means $$AP:PB = \lambda:\mu$$. The position vector $$\overrightarrow{OP}$$ is the sum of $$\overrightarrow{OA}$$ and $$\overrightarrow{AP}$$.

You can use the ratios of $$AP$$ over the whole of $$AB$$, times the the vector $$\overrightarrow{AB}$$ to work out $$\overrightarrow{AP}$$. So:​

​$$\boxed{\begin{aligned}\overrightarrow{OP}&=\overrightarrow{OA}+\frac{\lambda}{\lambda+\mu}\overrightarrow{AB}\\&=\overrightarrow{OA}+\frac{\lambda}{\lambda+\mu}(\overrightarrow{OB}-\overrightarrow{OA})\end{aligned}}$$​​

where $$\overrightarrow{AB}$$ is rewritten in terms of the position vectors of $$A$$ and $$B$$.

Example 1

The points $$A$$ and $$B$$ have position vectors $$\textbf{a}$$ and $$\textbf{b}$$, respectively. The point divides $$P$$ divides $$AB$$ in the ratio $$1:3$$. Find the position vector of $$P$$.

There are $$1+3=4$$ parts in the ratio in total. So, $$P$$ is $$\frac14$$ along $$AB$$.

​$$\overrightarrow{OP}=\overrightarrow{OA}+\frac{1}{4}\overrightarrow{AB}$$​

Or, using the formula and substituting for $$\lambda=1$$ and $$\mu=3$$:

$$\begin{aligned}\overrightarrow{OP}&=\overrightarrow{OA}+\frac{\lambda}{\lambda+\mu}\overrightarrow{AB}\\&=\overrightarrow{OA}+\frac1{1+3}\overrightarrow{AB}\\&=\overrightarrow{OA}+\frac14\overrightarrow{AB}\end{aligned}$$​​

Rewrite $$\overrightarrow{AB}$$ in terms of position vectors for $$A$$ and $$B$$:

$$\begin{aligned}\overrightarrow{OP}&=\overrightarrow{OA}+\frac14(\overrightarrow{OB}-\overrightarrow{OA})\\&=\textbf{a}+\frac14(\textbf{b}-\textbf{a})\\&=\underline{\frac34\textbf{a}+\frac14\textbf{b}}\end{aligned}$$​​

Comparing coefficients

For non-parallel vectors, $$\textbf{a}$$​ and $$\textbf{b}$$, you can compare the coefficients to solve geometric problems:

​$$p\textbf{a}+q\textbf{b}=r\textbf{a}+s\textbf{b}$$​​

​$$p=r, q=s$$​​

​

Example 2

​$$OABC$$ is a parallelogram, where $$\overrightarrow{OA}=\textbf{a}$$ and $$\overrightarrow{OC}=\textbf{c}$$​. The diagonals $$OB$$ and $$AC$$ intersect at point $$P$$. Prove that the diagonals bisect each other.

If the diagonals bisect each other, then $$P$$ is the midpoint of $$OB$$ and $$AC$$.

Use the fact that $$P$$ lies on both diagonals, i.e. you can find two different routes for $$P$$ from $$O$$, giving two different expressions for $$\overrightarrow{OP}$$​ in terms of $$\textbf{a}$$ and $$\textbf{c}$$. Consider $$OP$$ as a fraction $$\lambda$$ of $$OB$$, and the route $$O \rarr A\rarr P$$.

$$\begin{aligned}\overrightarrow{OP}&=\lambda\overrightarrow{OB}=\lambda(\overrightarrow{OC}+\overrightarrow{CB})=\lambda\textbf{(c}+\textbf{a})\end{aligned}$$​​

From the diagram, where $$\mu$$ is fraction of $$AC$$:

​$$\begin{aligned}\overrightarrow{OP}&=\overrightarrow{OA}+\overrightarrow{AP}=\overrightarrow{OA}+\mu\overrightarrow{AC}=\overrightarrow{OA}+\mu(-\overrightarrow{OA}+\overrightarrow{OC})=\textbf{a}+\mu(-\textbf{a}+\textbf{c})=(1-\mu)\textbf{a}+\mu\textbf{c}\end{aligned}$$​​

Comparing the coefficients of $$\textbf{a}$$ and $$\textbf{c}$$ from both expressions for $$\overrightarrow{OP}$$​ gives:

$$\lambda=1-\mu$$, $$\lambda=\mu$$​

​$$\underline{\lambda=\mu=\frac12}$$​​

So, $$P$$ is the midpoint of both diagonals, and so the diagonals bisect each other.

​

Triangles

You can apply your knowledge of triangle rules, using vectors.

Example 3

In triangle $$ABC$$, $$\overrightarrow{AB}=4\textbf{i}-\textbf{j}$$ and $$\overrightarrow{AC}=3\textbf{i}+3\textbf{j}$$. Find $$\angle BAC$$ in degrees to $$1\ d.p.$$​

Work out what information you need to find the angle.

One method to work out the angle is to use the cosine rule:

 $$\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{(|\overrightarrow{AB}|)^2+(|\overrightarrow{AC}|)^2-(|\overrightarrow{BC}|)^2}{2(|\overrightarrow{AB}|)(|\overrightarrow{AC}|)}$$​

To use this method, you need all the lengths of the triangle.

You can find the lengths of the triangle by working out the magnitude of the vectors. 

$$\begin{aligned}|\overrightarrow{AB}|&=\sqrt{4^2+(-1)^2}=\sqrt{17}\\|\overrightarrow{AC}|&=\sqrt{3^2+3^2}=\sqrt{18}\end{aligned}$$​​

​​

Note: Keeping answers in surd form will give a more accurate final answer, and they are easier to substitute for the cosine rule.

For the final length $$BC$$, find $$\overrightarrow{BC}$$:

$$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=\begin{pmatrix}3\\3\end{pmatrix}-\begin{pmatrix}4\\-1\end{pmatrix}=\begin{pmatrix}-1\\4\end{pmatrix}$$​​

​$$|\overrightarrow{BC}|=\sqrt{(-1)^2+4^2}=\sqrt{1+16}=\sqrt{17}$$​

Substitute values into the cosine rule:

$$\begin{aligned}\cos A&=\frac{b^2+c^2-a^2}{2bc}\\\\&=\frac{17+18-17}{2\times \sqrt{17}\times \sqrt{18}}\\\\A&=\underline{59.0}\degree\end{aligned}$$​​

​