Question

The two adjacent sides $OA,OB$ of parallelogram are $2\hat{i}+4\hat{j}-5\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$. The unit vectors along the diagonals of the parallelogram are given by

• A. $\frac{(3\hat{i}+6\hat{j}-2\hat{k})}{7}$
• B. $\frac{(-\hat{i}-2\hat{j}+8\hat{k})}{\sqrt{69}}$
• C. $\frac{(-3\hat{i}-6\hat{j}+2\hat{k})}{7}$
• D. $\frac{(\hat{i}+2\hat{j}-8\hat{k})}{\sqrt{69}}$

Answer 1

The correct option is A $\frac{(3\hat{i}+6\hat{j}-2\hat{k})}{7}$

Given that,

$\overrightarrow{OA}=2\hat{i}+4\hat{j}-5\hat{k}$

$\overrightarrow{OB}=\hat{i}+2\hat{j}+3\hat{k}$

Now, one diagonal of a parallelogram

$\overrightarrow{P}=\overrightarrow{A}+\overrightarrow{B}$

$\overrightarrow{P}=(2\hat{i}+4\hat{j}-5\hat{k})+(\hat{i}+2\hat{j}+3\hat{k})$

$\overrightarrow{P}=3\hat{i}+6\hat{j}-2\hat{k}$

Now, unit vector along the diagonal

$\hat{P}=\frac{\overrightarrow{P}}{|\overrightarrow{P}|}$

$\hat{P}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}$

$\hat{P}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$

Hence, the unit vectors along the diagonals of the parallelogram is $\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}$