Question
The vector $\bar{a}$ is directed due north and $|\bar{a}|=24$. The vector $\bar{b}$ is directed due west and $|\bar{b}|$
$=7$. Find $|\bar{a}+\bar{b}|$.
✓
Answer
Let $\overline{\mathrm{AB}}=\bar{a}_r \overline{\mathrm{BC}}=\bar{b}$
Then $\overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=\mathrm{a}+b$
Given: $|\bar{a}|=|\overline{\mathrm{AB}}|=1(\mathrm{AB})=24$ and
$|\bar{b}|=|\overline{\mathrm{BC}}|=\mid(\mathrm{BC})=7$
$\begin{aligned} & \therefore \angle A B C=90^{\circ} \\ & \therefore[I(A C)]^2=[I(A B)]^2+[I(B C)]^2\end{aligned}$
$\begin{aligned} & =(24)^2+(7)^2=625 \\ & \therefore|(\mathrm{AC})=25 \therefore| \overline{\mathrm{AC}} \mid=25 \\ & \therefore|\bar{a}+\bar{b}|=|\overline{\mathrm{AC}}|=25 .\end{aligned}$
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