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

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(A) Given $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$.We have $\vec{a} \cdot \vec{b} = (1)(1) + (2)(1) + (3)(-1) = 1 + 2

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$285$

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(A) Given $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$.We have $\vec{a} \cdot \vec{b} = (1)(1) + (2)(1) + (3)(-1) = 1 + 2 - 3 = 0$,so $\vec{a} \perp \vec{b}$.Also,$|\vec{b}| = \sqrt{1^2+1^2+(-1)^2} = \sqrt{3}$.Given $\vec{b} \cdot (\vec{a} 	imes \vec{c}) = 27$. This is the scalar triple product $[\vec{b}, \vec{a}, \vec{c}] = 27$.Using the vector triple product identity $\vec{b} 	imes (\vec{a} 	imes \vec{c}) = (\vec{b} \cdot \vec{c})\vec{a} - (\vec{b} \cdot \vec{a})\vec{c}$.Since $\vec{b} \cdot \vec{a} = 0$,we have $\vec{b} 	imes (\vec{a} 	imes \vec{c}) = (\vec{b} \cdot \vec{c})\vec{a}$.Given $\vec{b} \cdot \vec{c} = -\sqrt{3}|\vec{b}| = -\sqrt{3}(\sqrt{3}) = -3$.So,$\vec{b} 	imes (\vec{a} 	imes \vec{c}) = -3\vec{a}$.Taking the magnitude of both sides: $|\vec{b}| |\vec{a} 	imes \vec{c}| \sin 	heta = |-3\vec{a}| = 3|\vec{a}|$,where $	heta$ is the angle between $\vec{b}$ and $\vec{a} 	imes \vec{c}$.$|\vec{a}| = \sqrt{1^2+2^2+3^2} = \sqrt{14}$.So,$\sqrt{3} |\vec{a} 	imes \vec{c}| \sin 	heta = 3\sqrt{14}$.Also,$\vec{b} \cdot (\vec{a} 	imes \vec{c}) = |\vec{b}| |\vec{a} 	imes \vec{c}| \cos 	heta = 27$.$\sqrt{3} |\vec{a} 	imes \vec{c}| \cos 	heta = 27 \implies |\vec{a} 	imes \vec{c}| \cos 	heta = \frac{27}{\sqrt{3}} = 9\sqrt{3}$.Squaring and adding: $(|\vec{a} 	imes \vec{c}| \sin 	heta)^2 + (|\vec{a} 	imes \vec{c}| \cos 	heta)^2 = (\frac{3\sqrt{14}}{\sqrt{3}})^2 + (9\sqrt{3})^2$.$|\vec{a} 	imes \vec{c}|^2 = 3(14) + 81(3) = 42 + 243 = 285$.

Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=11$,$\vec{b} \cdot(\vec{a} \times \vec{c})=27$ and $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$,then $|\vec{a} \times \vec{c}|^2$ is equal to

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