Let vec{a}=4 hat{i}-hat{j}+hat{k},vec{b}=11 h | vedclass

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(B) Given $(\vec{a}+\vec{b}) 	imes \vec{c} = \vec{c} 	imes (-2 \vec{a}+3 \vec{b})$.This can be rewritten as $(\vec{a}+\vec{b}) 	imes \vec{c} + (-2

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$1618$

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(B) Given $(\vec{a}+\vec{b}) 	imes \vec{c} = \vec{c} 	imes (-2 \vec{a}+3 \vec{b})$.This can be rewritten as $(\vec{a}+\vec{b}) 	imes \vec{c} + (-2 \vec{a}+3 \vec{b}) 	imes \vec{c} = \vec{0}$.$(\vec{a}+\vec{b}-2 \vec{a}+3 \vec{b}) 	imes \vec{c} = \vec{0}$.$(4 \vec{b}-\vec{a}) 	imes \vec{c} = \vec{0}$.This implies $\vec{c}$ is parallel to $(4 \vec{b}-\vec{a})$.Let $\vec{c} = \lambda(4 \vec{b}-\vec{a})$.$4 \vec{b}-\vec{a} = 4(11 \hat{i}-\hat{j}+\hat{k}) - (4 \hat{i}-\hat{j}+\hat{k}) = (44-4)\hat{i} + (-4+1)\hat{j} + (4-1)\hat{k} = 40 \hat{i}-3 \hat{j}+3 \hat{k}$.So,$\vec{c} = \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k})$.Given $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$.$2 \vec{a}+3 \vec{b} = 2(4 \hat{i}-\hat{j}+\hat{k}) + 3(11 \hat{i}-\hat{j}+\hat{k}) = (8+33)\hat{i} + (-2-3)\hat{j} + (2+3)\hat{k} = 41 \hat{i}-5 \hat{j}+5 \hat{k}$.$(41 \hat{i}-5 \hat{j}+5 \hat{k}) \cdot \lambda(40 \hat{i}-3 \hat{j}+3 \hat{k}) = 1670$.$\lambda(41 	imes 40 + (-5) 	imes (-3) + 5 	imes 3) = 1670$.$\lambda(1640 + 15 + 15) = 1670$.$1670 \lambda = 1670 \Rightarrow \lambda = 1$.Thus,$\vec{c} = 40 \hat{i}-3 \hat{j}+3 \hat{k}$.$|\vec{c}|^2 = 40^2 + (-3)^2 + 3^2 = 1600 + 9 + 9 = 1618$.

Let $\vec{a}=4 \hat{i}-\hat{j}+\hat{k}$,$\vec{b}=11 \hat{i}-\hat{j}+\hat{k}$,and $\vec{c}$ be a vector such that $(\vec{a}+\vec{b}) \times \vec{c} = \vec{c} \times (-2 \vec{a}+3 \vec{b})$. If $(2 \vec{a}+3 \vec{b}) \cdot \vec{c} = 1670$,then $|\vec{c}|^2$ is equal to:

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