If vec{a}=2 hat{i}+hat{j}+2 hat{k}, then the | 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

$18$

Vedclass · Answer

(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 have $\hat{i} 	imes (\vec{a} 	imes \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} = \vec{a} - a_x \hat{i}$.Similarly,$\hat{j} 	imes (\vec{a} 	imes \hat{j}) = \vec{a} - a_y \hat{j}$ and $\hat{k} 	imes (\vec{a} 	imes \hat{k}) = \vec{a} - a_z \hat{k}$.Given $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$,we have $|\vec{a}|^2 = 2^2 + 1^2 + 2^2 = 4 + 1 + 4 = 9$.The expression is $|\vec{a} - a_x \hat{i}|^2 + |\vec{a} - a_y \hat{j}|^2 + |\vec{a} - a_z \hat{k}|^2$.Expanding each term: $|\vec{a}|^2 + a_x^2 - 2a_x^2 = |\vec{a}|^2 - a_x^2$,$|\vec{a}|^2 - a_y^2$,and $|\vec{a}|^2 - a_z^2$.Summing these: $3|\vec{a}|^2 - (a_x^2 + a_y^2 + a_z^2) = 3|\vec{a}|^2 - |\vec{a}|^2 = 2|\vec{a}|^2$.Substituting $|\vec{a}|^2 = 9$,we get $2 	imes 9 = 18$.

If $\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k},$ then the value of $|\hat{i} \times(\vec{a} \times \hat{i})|^{2}+|\hat{j} \times(\vec{a} \times \hat{j})|^{2}+|\hat{k} \times(\vec{a} \times \hat{k})|^{2}$ is equal to

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