Let vec{a}=2 hat{i}-hat{j}+5 hat{k} and vec{b | vedclass

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(B) Given $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$.Using the vector triple product formula $(\vec{u}

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$5$

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(B) Given $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$.Using the vector triple product formula $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$,we have:$((\vec{a} 	imes \vec{b}) 	imes \hat{i}) \cdot \hat{k} = ((\vec{a} \cdot \hat{i})\vec{b} - (\vec{b} \cdot \hat{i})\vec{a}) \cdot \hat{k} = \frac{23}{2}$.Since $\vec{a} \cdot \hat{i} = 2$ and $\vec{b} \cdot \hat{i} = \alpha$,the expression becomes:$(2\vec{b} - \alpha\vec{a}) \cdot \hat{k} = 2(\vec{b} \cdot \hat{k}) - \alpha(\vec{a} \cdot \hat{k}) = \frac{23}{2}$.Given $\vec{b} \cdot \hat{k} = 2$ and $\vec{a} \cdot \hat{k} = 5$,we get:$2(2) - \alpha(5) = \frac{23}{2} \implies 4 - 5\alpha = \frac{23}{2} \implies 5\alpha = 4 - \frac{23}{2} = -\frac{15}{2} \implies \alpha = -\frac{3}{2}$.Now,calculate $\vec{b} 	imes 2\hat{j} = 2(\vec{b} 	imes \hat{j})$:$\vec{b} 	imes \hat{j} = (\alpha \hat{i} + \beta \hat{j} + 2 \hat{k}) 	imes \hat{j} = \alpha(\hat{i} 	imes \hat{j}) + \beta(\hat{j} 	imes \hat{j}) + 2(\hat{k} 	imes \hat{j}) = \alpha \hat{k} - 2 \hat{i}$.So,$\vec{b} 	imes 2\hat{j} = 2(-\alpha \hat{k} + 2 \hat{i}) = 4\hat{i} - 2\alpha \hat{k}$.$|\vec{b} 	imes 2\hat{j}| = \sqrt{4^2 + (-2\alpha)^2} = \sqrt{16 + 4\alpha^2} = \sqrt{16 + 4(-\frac{3}{2})^2} = \sqrt{16 + 4(\frac{9}{4})} = \sqrt{16 + 9} = \sqrt{25} = 5$.

Let $\vec{a}=2 \hat{i}-\hat{j}+5 \hat{k}$ and $\vec{b}=\alpha \hat{i}+\beta \hat{j}+2 \hat{k}$. If $((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k}=\frac{23}{2}$,then $|\vec{b} \times 2 \hat{j}|$ is equal to.

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