The unit vectors orthogonal to 3 hat{i}+2 hat | vedclass

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(B) Let the required vector be $\vec{v} = a(2 \hat{i}+\hat{j}+\hat{k}) + b(\hat{i}-\hat{j}+\hat{k})$.This simplifies to $\vec{v} = (2a+b)\hat{i} + (a-

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$\pm \frac{1}{\sqrt{10}}(3 \hat{j}-\hat{k})$

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(B) Let the required vector be $\vec{v} = a(2 \hat{i}+\hat{j}+\hat{k}) + b(\hat{i}-\hat{j}+\hat{k})$.This simplifies to $\vec{v} = (2a+b)\hat{i} + (a-b)\hat{j} + (a+b)\hat{k}$.Since $\vec{v}$ is orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$,their dot product is zero:$3(2a+b) + 2(a-b) + 6(a+b) = 0$.$6a + 3b + 2a - 2b + 6a + 6b = 0$.$14a + 7b = 0 \implies b = -2a$.Substituting $b = -2a$ into the expression for $\vec{v}$:$\vec{v} = (2a - 2a)\hat{i} + (a - (-2a))\hat{j} + (a + (-2a))\hat{k} = 0\hat{i} + 3a\hat{j} - a\hat{k} = a(3\hat{j} - \hat{k})$.The unit vector is $\pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{a(3\hat{j} - \hat{k})}{|a|\sqrt{3^2 + (-1)^2}} = \pm \frac{3\hat{j} - \hat{k}}{\sqrt{10}}$.

The unit vectors orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are

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