Question

If the diagonals of a parallelogram are given $3\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}+4\hat{k}$, then the lengths of its sides are

• A. $\sqrt{8},\sqrt{10}$
• B. $\sqrt{6},\sqrt{14}$
• C. $\sqrt{5},\sqrt{12}$
• D. $None$

Answer 1

The correct option is B $\sqrt{6},\sqrt{14}$

Let a, b be the sides of the parallelogram. Diagonals are a+b, a-b.
$a+b=3\hat{i}+\hat{j}-2\hat{k},a-b=\hat{i}-3\hat{j}+4\hat{k}\Rightarrow a=\frac{1}{2}(4\hat{i}-2\hat{j}+2\hat{k})=2\hat{i}-\hat{j}+\hat{k}b=\frac{1}{2}(2\hat{i}+4\hat{j}-6\hat{k})=\hat{i}+2\hat{j}-3\hat{k}|a|=\sqrt{4+1+1}=\sqrt{6},|b|=\sqrt{1+4+9}=\sqrt{14}$
$\therefore$ Lengths of the sides are $\sqrt{6},\sqrt{14}$.