Question
If $\overline{\mathrm{AB}}=2 \hat{i}-4 \hat{j}+7 \hat{k}$ and initial point $\mathrm{A} \equiv(1,5,0)$. Find the terminal point $\mathrm{B}$.
Given: $\mathrm{A}=(1,5,0) \therefore \bar{a}=\hat{i}+5 \hat{j}$
Now, $\overline{\mathrm{AB}}=2 \hat{i}-4 \hat{j}+7 \hat{k}$
$\begin{aligned} & \therefore \bar{b}-\bar{a}=2 \hat{i}-4 \hat{j}+7 \hat{k} \\ & \therefore \bar{b}=(2 \hat{i}-4 \hat{j}+7 \hat{k})+\bar{a} \\ & =(2 \hat{i}-4 \hat{j}+7 \hat{k})+(\hat{i}+5 \hat{j}) \\ & =3 \hat{i}+\hat{j}+7 \hat{k}\end{aligned}$
Hence, the terminal point B = (3, 1, 7).
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