Question

If the three successive vertices of a parallelogram have the position vectors as, $A(-3,-2,0);B(3,-3,1)$ and $C(5,0,2).$ Then find :
a vector having the same direction as that of $\overrightarrow{AB}$ but magnitude equal to $\overrightarrow{AC}$

Answer 1

$A(-3,-2,0)B(3,-3,1)C(5,0,2)$

$\overrightarrow{OA}=-3\hat{i}-2\hat{j}-0\overrightarrow{OB}=3\hat{i}-3\hat{j}+\hat{k}\overrightarrow{OC}=5\hat{i}+2\hat{k}$

$\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=5\hat{i}+2\hat{k}-(-3\hat{i}-2\hat{j})$

$\overrightarrow{AC}=8\hat{i}+2\hat{j}+2\hat{k}$

$|\overrightarrow{AC}|=\sqrt{8^2+7^2+2^2}=\sqrt{72}$

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(3\hat{i}-3\hat{j}-\hat{k})-(-3\hat{i}-\hat{j})$

$\overrightarrow{AB}=6\hat{i}-\hat{j}+\hat{k}$

Vector in direction of $\overrightarrow{AB}=k(6\hat{i}\hat{j}+\hat{k})$

it has magnitude $\sqrt{72}$

$\Rightarrow \sqrt{72}=\sqrt{(6k{)}^2+(-k{)}^2+k^2}$

$\Rightarrow \sqrt{72}=k\sqrt{38}$

$k=\sqrt{\frac{72}{38}}=\sqrt{\frac{36}{19}}=\frac{6}{\sqrt{19}}$

$\therefore$ Vector direction of $vecAB$ which magnitude of $\overrightarrow{AC}$

$\Rightarrow \frac{6}{\sqrt{19}}(6\hat{i}-\hat{j}+\hat{k})$