Q118 / 180 · KCET 2021 · Mathematics
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
- Correct answerA$\sqrt{29}$
- B$\sqrt{14}$
- C$3 \sqrt{5}$
- D$4 \sqrt{3}$
Solution
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 $\quad \mathbf{b}=\frac{\mathbf{d}_{1}-\mathbf{d}_{2}}{2}=\frac{4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{2}$ $=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ $\Rightarrow \quad|\mathbf{b}|=\sqrt{4+16+9}=\sqrt{29}$
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 $\quad \mathbf{b}=\frac{\mathbf{d}_{1}-\mathbf{d}_{2}}{2}=\frac{4 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}}{2}$ $=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ $\Rightarrow \quad|\mathbf{b}|=\sqrt{4+16+9}=\sqrt{29}$
Hence, the correct answer is option: 'A'.