Let vec{a}=2 hat{i}+3 hat{j}+4 hat{k}, vec{b} | vedclass

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(D) Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$,$\vec{d}$ must be parallel to $\vec{b} 	imes \vec{c}$.Let $\vec{d} = \lambda(\ve

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(D) Since $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$,$\vec{d}$ must be parallel to $\vec{b} 	imes \vec{c}$.Let $\vec{d} = \lambda(\vec{b} 	imes \vec{c})$.First,calculate $\vec{b} 	imes \vec{c}$:$\vec{b} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -2 & -2 \ -1 & 4 & 3 \end{vmatrix} = \hat{i}(-6 - (-8)) - \hat{j}(6 - 2) + \hat{k}(8 - 2) = 2\hat{i} - 4\hat{j} + 6\hat{k}$.So,$\vec{d} = \lambda(2\hat{i} - 4\hat{j} + 6\hat{k})$.Given $\vec{a} \cdot \vec{d} = 18$:$(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda(2\hat{i} - 4\hat{j} + 6\hat{k}) = 18$$\lambda(4 - 12 + 24) = 18 \implies 16\lambda = 18 \implies \lambda = \frac{18}{16} = \frac{9}{8}$.Thus,$\vec{d} = \frac{9}{8}(2\hat{i} - 4\hat{j} + 6\hat{k}) = \frac{9}{4}\hat{i} - \frac{9}{2}\hat{j} + \frac{27}{4}\hat{k}$.Now,calculate $\vec{a} 	imes \vec{d} = \vec{a} 	imes (\lambda(\vec{b} 	imes \vec{c})) = \lambda(\vec{a} 	imes (\vec{b} 	imes \vec{c}))$.Using the vector triple product formula $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = \vec{b}(\vec{a} \cdot \vec{c}) - \vec{c}(\vec{a} \cdot \vec{b})$:$\vec{a} \cdot \vec{c} = (2)(-1) + (3)(4) + (4)(3) = -2 + 12 + 12 = 22$.$\vec{a} \cdot \vec{b} = (2)(2) + (3)(-2) + (4)(-2) = 4 - 6 - 8 = -10$.$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = 22(2\hat{i} - 2\hat{j} - 2\hat{k}) - (-10)(-\hat{i} + 4\hat{j} + 3\hat{k}) = (44\hat{i} - 44\hat{j} - 44\hat{k}) - (10\hat{i} - 40\hat{j} - 30\hat{k}) = 34\hat{i} - 4\hat{j} - 14\hat{k}$.$\vec{a} 	imes \vec{d} = \frac{9}{8}(34\hat{i} - 4\hat{j} - 14\hat{k}) = \frac{9}{4}(17\hat{i} - 2\hat{j} - 7\hat{k})$.$|\vec{a} 	imes \vec{d}|^2 = (\frac{9}{4})^2 (17^2 + (-2)^2 + (-7)^2) = \frac{81}{16} (289 + 4 + 49) = \frac{81}{16} (342) = \frac{81 	imes 171}{8} = 1732.875$.

Let $\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}-2 \hat{k}$ and $\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}$. If $\vec{d}$ is a vector perpendicular to both $\vec{b}$ and $\vec{c}$ and $\vec{a} \cdot \vec{d}=18$,then $|\vec{a} \times \vec{d}|^2$ is equal to $..........$.

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