If overrightarrow{a} = hat{i} + hat{j} + hat{ | vedclass

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(A) Let the diagonals of the parallelogram be $\overrightarrow{d_1} = \overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{d_2} = \overrighta

Vedclass · Accepted Answer

$4\sqrt{6}$ sq units

Vedclass · Answer

(A) Let the diagonals of the parallelogram be $\overrightarrow{d_1} = \overrightarrow{a} + \overrightarrow{b}$ and $\overrightarrow{d_2} = \overrightarrow{b} + \overrightarrow{c}$.$\overrightarrow{d_1} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 3\hat{j} + 5\hat{k}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$$\overrightarrow{d_2} = (\hat{i} + 3\hat{j} + 5\hat{k}) + (7\hat{i} + 9\hat{j} + 11\hat{k}) = 8\hat{i} + 12\hat{j} + 16\hat{k}$The area of a parallelogram with diagonals $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$ is given by $\frac{1}{2} |\overrightarrow{d_1} 	imes \overrightarrow{d_2}|$.$\overrightarrow{d_1} 	imes \overrightarrow{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 4 & 6 \ 8 & 12 & 16 \end{vmatrix} = \hat{i}(64 - 72) - \hat{j}(32 - 48) + \hat{k}(24 - 32) = -8\hat{i} + 16\hat{j} - 8\hat{k}$Magnitude $|\overrightarrow{d_1} 	imes \overrightarrow{d_2}| = \sqrt{(-8)^2 + 16^2 + (-8)^2} = \sqrt{64 + 256 + 64} = \sqrt{384} = \sqrt{64 	imes 6} = 8\sqrt{6}$Area $= \frac{1}{2} 	imes 8\sqrt{6} = 4\sqrt{6}$ sq units.

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