If vec{a} = hat{i} + 2hat{j} + 3hat{k} and ve | vedclass

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(D) Using the vector triple product identity $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w

Vedclass · Accepted Answer

$2\sqrt{14}$

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(D) Using the vector triple product identity $\vec{u} 	imes (\vec{v} 	imes \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$,we evaluate each term: $\hat{i} 	imes (\vec{a} 	imes \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} = 1(\vec{a}) - (1)\hat{i} = (\hat{i} + 2\hat{j} + 3\hat{k}) - \hat{i} = 2\hat{j} + 3\hat{k}$ $\hat{j} 	imes (\vec{a} 	imes \hat{j}) = (\hat{j} \cdot \hat{j})\vec{a} - (\hat{j} \cdot \vec{a})\hat{j} = 1(\vec{a}) - (2)\hat{j} = (\hat{i} + 2\hat{j} + 3\hat{k}) - 2\hat{j} = \hat{i} + 3\hat{k}$ $\hat{k} 	imes (\vec{a} 	imes \hat{k}) = (\hat{k} \cdot \hat{k})\vec{a} - (\hat{k} \cdot \vec{a})\hat{k} = 1(\vec{a}) - (3)\hat{k} = (\hat{i} + 2\hat{j} + 3\hat{k}) - 3\hat{k} = \hat{i} + 2\hat{j}$ Summing these results: $\vec{b} = (2\hat{j} + 3\hat{k}) + (\hat{i} + 3\hat{k}) + (\hat{i} + 2\hat{j}) = 2\hat{i} + 4\hat{j} + 6\hat{k}$ Finally,the magnitude is: $|\vec{b}| = \sqrt{2^2 + 4^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = \sqrt{4 	imes 14} = 2\sqrt{14}$

If $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} \times (\vec{a} \times \hat{i}) + \hat{j} \times (\vec{a} \times \hat{j}) + \hat{k} \times (\vec{a} \times \hat{k})$,then $|\vec{b}|$ is

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