The diagonals of a parallelogram are the vectors \(3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\). and \(-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}\). Then the length of the shorter side of parallelogram is

Let \(\mathbf{a}\) and \(\mathbf{b}\) be the length of the sides of the parallelogram and \(\mathbf{d}_{1}, \mathbf{d}_{2}\) be the length of diagonals.



Given, \(\begin{aligned} \mathbf{d}_{1} &=3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ \mathbf{d}_{2} &=-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-8 \hat{\mathbf{k}} \end{aligned}\)
\(\therefore \quad \mathbf{a}=\frac{\mathbf{d}_{1}+\mathbf{d}_{2}}{2}=\frac{2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}}{2}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}\)
\(\Rightarrow \quad|\mathbf{a}|=\sqrt{1+4+25}=\sqrt{30}\)
and \(\quad \mathbf{b}=\frac{\mathbf{d}_{1}-\mathbf{d}_{2}}{2}=\frac{4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{2}\)
\(=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
\(\Rightarrow \quad|\mathbf{b}|=\sqrt{4+16+9}=\sqrt{29}\)