Question

The sides of a parallelogram are $2\hat{i}+4\hat{j}-5\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$. The unit vector parallel to one of the diagonals is

• A. $\frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k})$
• B. $\frac{1}{7}(3\hat{i}-6\hat{j}-2\hat{k})$
• C. $\frac{1}{\sqrt{69}}(\hat{i}+2\hat{j}+8\hat{k})$
• D. $\frac{1}{\sqrt{69}}-(\hat{i}-2\hat{j}+8\hat{k})$

Answer 1

The correct option is A $\frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k})$

Adjacent sides of a parallelogram are given as:

$\overrightarrow{a}=2\hat{i}+4\hat{j}-5\hat{k}and\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$

Then, the diagonal of a parallelogram is given by

$\begin{array}{c}\overrightarrow{a}+\overrightarrow{b}\\ \overrightarrow{a}+\overrightarrow{b}=\left(2+1\right)\hat{i}+\left(4+2\right)\hat{j}+\left(-5+3\right)\hat{k}=3\hat{i}+6\hat{j}-2\hat{k}\\ \frac{\overrightarrow{a}+\overrightarrow{b}}{\left|\overrightarrow{a}\right|+\left|\overrightarrow{b}\right|}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{9+36+4}}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7}=\frac{3}{7}\hat{i}+\frac{6}{7}\hat{j}-\frac{2}{7}\hat{k}\\ =\frac{1}{7}\left(3\hat{i}+6\hat{j}-2\hat{k}\right)\end{array}$