If vec{a} = hat{i} + 2hat{j} - 2hat{k},vec{b} | vedclass

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(A) Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c

Vedclass · Accepted Answer

$20\hat{i} - 3\hat{j} + 7\hat{k}$

Vedclass · Answer

(A) Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.First,calculate the dot products:$\vec{a} \cdot \vec{c} = (1)(1) + (2)(3) + (-2)(-1) = 1 + 6 + 2 = 9$.$\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (-2)(1) = 2 - 2 - 2 = -2$.Now,substitute these values into the formula:$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = 9(2\hat{i} - \hat{j} + \hat{k}) - (-2)(\hat{i} + 3\hat{j} - \hat{k})$.$= (18\hat{i} - 9\hat{j} + 9\hat{k}) + (2\hat{i} + 6\hat{j} - 2\hat{k})$.$= (18+2)\hat{i} + (-9+6)\hat{j} + (9-2)\hat{k} = 20\hat{i} - 3\hat{j} + 7\hat{k}$.

If $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$,$\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$,and $\vec{c} = \hat{i} + 3\hat{j} - \hat{k}$,then find $\vec{a} \times (\vec{b} \times \vec{c})$.

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Statement $(A)$ : If $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,then $\vec{a} \times (\vec{b} \times \vec{c}) = 0$.
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