Let vec{a}=2 hat{i}+7 hat{j}-hat{k}, vec{b}=3 | vedclass

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(C) Given $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}$,$\vec{b}=3 \hat{i}+5 \hat{k}$,and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$.Since $\vec{d}$ is perpendicular

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$44$

Vedclass · Answer

(C) Given $\vec{a}=2 \hat{i}+7 \hat{j}-\hat{k}$,$\vec{b}=3 \hat{i}+5 \hat{k}$,and $\vec{c}=\hat{i}-\hat{j}+2 \hat{k}$.Since $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$,$\vec{d} = \lambda(\vec{a} 	imes \vec{b})$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 7 & -1 \ 3 & 0 & 5 \end{vmatrix} = \hat{i}(35-0) - \hat{j}(10+3) + \hat{k}(0-21) = 35\hat{i} - 13\hat{j} - 21\hat{k}$.So,$\vec{d} = \lambda(35\hat{i} - 13\hat{j} - 21\hat{k})$.Given $\vec{c} \cdot \vec{d} = 12$,we have $\lambda(\hat{i}-\hat{j}+2\hat{k}) \cdot (35\hat{i} - 13\hat{j} - 21\hat{k}) = 12$.$\lambda(35 + 13 - 42) = 12 \implies 6\lambda = 12 \implies \lambda = 2$.Thus,$\vec{d} = 2(35\hat{i} - 13\hat{j} - 21\hat{k}) = 70\hat{i} - 26\hat{j} - 42\hat{k}$.We need to find $(-\hat{i}+\hat{j}-\hat{k}) \cdot (\vec{c} 	imes \vec{d})$.Since $\vec{d} = \lambda(\vec{a} 	imes \vec{b})$,$\vec{c} 	imes \vec{d} = \lambda(\vec{c} 	imes (\vec{a} 	imes \vec{b})) = \lambda((\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b})$.Alternatively,calculate the scalar triple product using the determinant:$(-\hat{i}+\hat{j}-\hat{k}) \cdot (\vec{c} 	imes \vec{d}) = \begin{vmatrix} -1 & 1 & -1 \ 1 & -1 & 2 \ 70 & -26 & -42 \end{vmatrix} = -1(42 - (-52)) - 1(-42 - 140) - 1(-26 - (-70)) = -1(94) - 1(-182) - 1(44) = -94 + 182 - 44 = 44$.

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