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Question

(A) vector perpendicular to both $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is given by their cross product: $(\vec{a} + \vec{b}) 	imes (\vec{a}

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k}$

Vedclass · Answer

(A) vector perpendicular to both $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is given by their cross product: $(\vec{a} + \vec{b}) 	imes (\vec{a} - \vec{b})$.Expanding this, we get: $\vec{a} 	imes \vec{a} - \vec{a} 	imes \vec{b} + \vec{b} 	imes \vec{a} - \vec{b} 	imes \vec{b} = 0 - (\vec{a} 	imes \vec{b}) - (\vec{a} 	imes \vec{b}) - 0 = -2(\vec{a} 	imes \vec{b})$.First, calculate $\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-2) - \hat{j}(3-1) + \hat{k}(2-1) = \hat{i} - 2\hat{j} + \hat{k}$.Thus, the vector perpendicular to both is $-2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$.The unit vector is $\pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{(-2)^2 + 4^2 + (-2)^2}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{4 + 16 + 4}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{\sqrt{24}} = \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}}$.This simplifies to $\pm (-\frac{1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k})$. Option $A$ matches this result.

$A$ unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is . . . . . . where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

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