If the diagonals of a parallelogram are given by 3i+j-2k and i-3j+4k, then the lengths of its sides are

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \overrightarrow{{\mathrm{d}}_1}=3\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ \overrightarrow{{\mathrm{d}}_2}=\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{Length} \mathrm{of} \mathrm{sides} \mathrm{are}\frac{ \left|\overrightarrow{{\mathrm{d}}_1}+\overrightarrow{{\mathrm{d}}_2}\right|}{2} \mathrm{and} \frac{\left|\overrightarrow{{\mathrm{d}}_1}-\overrightarrow{{\mathrm{d}}_2}\right|}{2} \\ \frac{ \left|\overrightarrow{{\mathrm{d}}_1}+\overrightarrow{{\mathrm{d}}_2}\right|}{2}=\frac{\left|3\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}+\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}\right|}{2} \\ =\frac{\left|4\hat{\mathrm{i}}-2\hat{\mathrm{j}}+2\hat{\mathrm{k}}\right|}{2}=\frac{\sqrt{16+4+4}}{2}=\frac{\sqrt{24}}{2}=\frac{2\sqrt{6}}{2}=\sqrt{\mathbf{6}} \\ \frac{\left|\overrightarrow{{\mathrm{d}}_1}-\overrightarrow{{\mathrm{d}}_2}\right|}{2}=\frac{\left|3\hat{\mathrm{i}}+\hat{\mathrm{j}}-2\hat{\mathrm{k}}-\hat{\mathrm{i}}+3\hat{\mathrm{j}}-4\hat{\mathrm{k}}\right|}{2} \\ =\frac{\left|2\hat{\mathrm{i}}+4\hat{\mathrm{j}}-6\hat{\mathrm{k}}\right|}{2}=\frac{\sqrt{4+16+36}}{2}=\frac{\sqrt{56}}{2}=\frac{2\sqrt{14}}{2}=\sqrt{\mathbf{14}} \end{gathered}$$

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