Let vec{a}=hat{i}+2hat{j}+hat{k} and vec{b}=2 | vedclass

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(D) Let the vector $\vec{p}$ lie in the plane of $\vec{a}$ and $\vec{b}$. Then $\vec{p} = \vec{a} + \lambda \vec{b}$.Since $\vec{p}$ is perpendicular

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

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(D) Let the vector $\vec{p}$ lie in the plane of $\vec{a}$ and $\vec{b}$. Then $\vec{p} = \vec{a} + \lambda \vec{b}$.Since $\vec{p}$ is perpendicular to $\vec{a}$,we have $\vec{p} \cdot \vec{a} = 0$.$(\vec{a} + \lambda \vec{b}) \cdot \vec{a} = 0 \Rightarrow \vec{a} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$.Given $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$.$\vec{a} \cdot \vec{a} = 1^2 + 2^2 + 1^2 = 6$.$\vec{b} \cdot \vec{a} = (2)(1) + (1)(2) + (-1)(1) = 2 + 2 - 1 = 3$.Substituting these values: $6 + \lambda(3) = 0 \Rightarrow \lambda = -2$.Thus,$\vec{p} = (\hat{i} + 2\hat{j} + \hat{k}) - 2(2\hat{i} + \hat{j} - \hat{k}) = -3\hat{i} + 3\hat{k}$.The unit vector $\hat{c}$ is $\pm \frac{\vec{p}}{|\vec{p}|} = \pm \frac{-3\hat{i} + 3\hat{k}}{\sqrt{(-3)^2 + 3^2}} = \pm \frac{-3\hat{i} + 3\hat{k}}{3\sqrt{2}} = \pm \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$.Comparing with the options,the correct vector is $\frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$.

Let $\vec{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$. Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :

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