The unit vector which is orthogonal to the ve | vedclass

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(C) Let $\vec{a}=5 \hat{i}+2 \hat{j}+6 \hat{k}$,$\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$,and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$.Since the required vector is

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$\frac{3 \hat{j}-\hat{k}}{\sqrt{10}}$

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(C) Let $\vec{a}=5 \hat{i}+2 \hat{j}+6 \hat{k}$,$\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$,and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$.Since the required vector is orthogonal to $\vec{a}$ and coplanar with $\vec{b}$ and $\vec{c}$,it must be parallel to $\vec{a} 	imes (\vec{b} 	imes \vec{c})$.Using the vector triple product formula: $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}$.Calculate the dot products:$\vec{a} \cdot \vec{c} = (5)(1) + (2)(-1) + (6)(1) = 5 - 2 + 6 = 9$.$\vec{a} \cdot \vec{b} = (5)(2) + (2)(1) + (6)(1) = 10 + 2 + 6 = 18$.Substitute these into the formula:$\vec{a} 	imes (\vec{b} 	imes \vec{c}) = 9(2 \hat{i} + \hat{j} + \hat{k}) - 18(\hat{i} - \hat{j} + \hat{k}) = (18-18)\hat{i} + (9+18)\hat{j} + (9-18)\hat{k} = 27 \hat{j} - 9 \hat{k}$.The magnitude is $|27 \hat{j} - 9 \hat{k}| = \sqrt{27^2 + (-9)^2} = \sqrt{729 + 81} = \sqrt{810} = 9 \sqrt{10}$.The unit vector is $\pm \frac{27 \hat{j} - 9 \hat{k}}{9 \sqrt{10}} = \pm \frac{3 \hat{j} - \hat{k}}{\sqrt{10}}$.Comparing with the options,$\frac{3 \hat{j} - \hat{k}}{\sqrt{10}}$ is the correct choice.

The unit vector which is orthogonal to the vector $5 \hat{i}+2 \hat{j}+6 \hat{k}$ and is coplanar with the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

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