Question

The sides of 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

Answer: (A), (D)

The correct options are
A $\frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k})$
D $\frac{1}{\sqrt{69}}(-\hat{i}-2\hat{j}+8\hat{k})$

Let $\overrightarrow{a}=2\hat{i}+4\hat{j}-5\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}+3\hat{k}$.
Then the diagonals of the parallelogram are
$\overrightarrow{p}=\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{q}=\overrightarrow{b}-\overrightarrow{a}.$
i.e., $\overrightarrow{p}=3\hat{i}+6\hat{j}-2\hat{k},\overrightarrow{q}=-\hat{i}-2\hat{j}+8\hat{k}$
So, unit vectors along the diagonals are
$\frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k})$ and $\frac{1}{\sqrt{69}}(-\hat{i}-2\hat{j}+8\hat{k})$.