A unit vector coplanar with hat{i}+hat{j}+hat | vedclass

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(A) Let $\vec{a} = \hat{i}+\hat{j}-\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = 2\hat{i}+\hat{j}+\hat{k}$.Since the required vector is

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$+\frac{1}{\sqrt{2}}(-\hat{j}-\hat{k})$

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(A) Let $\vec{a} = \hat{i}+\hat{j}-\hat{k}$,$\vec{b} = \hat{i}+\hat{j}+\hat{k}$,and $\vec{c} = 2\hat{i}+\hat{j}+\hat{k}$.Since the required vector is coplanar with $\vec{b}$ and $\vec{c}$,it must be of the form $\vec{v} = \vec{a} 	imes (\vec{b} 	imes \vec{c})$.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:$\vec{a} \cdot \vec{c} = (1)(2) + (1)(1) + (-1)(1) = 2 + 1 - 1 = 2$.$\vec{a} \cdot \vec{b} = (1)(1) + (1)(1) + (-1)(1) = 1 + 1 - 1 = 1$.Now,substitute these into the formula:$\vec{v} = 2(\hat{i}+\hat{j}+\hat{k}) - 1(2\hat{i}+\hat{j}+\hat{k}) = (2\hat{i}+2\hat{j}+2\hat{k}) - (2\hat{i}+\hat{j}+\hat{k}) = \hat{j}+\hat{k}$.The magnitude of this vector is $|\vec{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.The unit vector is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{\hat{j}+\hat{k}}{\sqrt{2}}$.Comparing with the given options,the correct choice is $A$ (considering the negative sign).

$A$ unit vector coplanar with $\hat{i}+\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ and perpendicular to $\hat{i}+\hat{j}-\hat{k}$ is

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