Question

The sides of a parallelogram are given by the vectors <2,4,-5> and <1,2,3> , then 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. $\frac{1}{7}(-3\hat{i}+6\hat{j}-2\hat{k})$
• D. $\frac{1}{7}(3\hat{i}+6\hat{j}+2\hat{k})$

Answer 1

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

Let $\overrightarrow{a}$= <2,4,-5> , $\overrightarrow{b}$= <1,2,3> . Then the diagonals of the parallelogram are $\overrightarrow{p}=\overrightarrow{a}+\overrightarrow{b},\overrightarrow{q}=\overrightarrow{b}-\overrightarrow{a}$
$\Rightarrow \overrightarrow{p}$= <3,6,-2> , $\overrightarrow{q}$= <-1,-2,8>
So, unit vectors along the diagonals are $\frac{1}{7}$ <3,6,-2> and $\frac{1}{\sqrt{69}}$ <-1,-2,8>