Let overrightarrow{a} = hat{i} + 2hat{j} + ha | vedclass

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(D) Given $\overrightarrow{b} 	imes \overrightarrow{d} = \overrightarrow{c} 	imes \overrightarrow{d}$,we have $(\overrightarrow{b} - \overrightarrow

Vedclass · Accepted Answer

$128$

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(D) Given $\overrightarrow{b} 	imes \overrightarrow{d} = \overrightarrow{c} 	imes \overrightarrow{d}$,we have $(\overrightarrow{b} - \overrightarrow{c}) 	imes \overrightarrow{d} = \overrightarrow{0}$.This implies $\overrightarrow{d} = \lambda(\overrightarrow{b} - \overrightarrow{c})$ for some scalar $\lambda$.Calculating $\overrightarrow{b} - \overrightarrow{c} = (3-2)\hat{i} + (-3 - (-1))\hat{j} + (3-2)\hat{k} = \hat{i} - 2\hat{j} + \hat{k}$.So,$\overrightarrow{d} = \lambda(\hat{i} - 2\hat{j} + \hat{k})$.Given $\overrightarrow{a} \cdot \overrightarrow{d} = 4$,we have $(\hat{i} + 2\hat{j} + \hat{k}) \cdot \lambda(\hat{i} - 2\hat{j} + \hat{k}) = 4$.$\lambda(1 - 4 + 1) = 4 \Rightarrow -2\lambda = 4 \Rightarrow \lambda = -2$.Thus,$\overrightarrow{d} = -2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.Now,$\overrightarrow{a} 	imes \overrightarrow{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 1 \ -2 & 4 & -2 \end{vmatrix} = \hat{i}(-4 - 4) - \hat{j}(-2 - (-2)) + \hat{k}(4 - (-4)) = -8\hat{i} + 8\hat{k}$.Then $|\overrightarrow{a} 	imes \overrightarrow{d}|^2 = (-8)^2 + 0^2 + 8^2 = 64 + 64 = 128$.

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