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

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(C) Let $\vec{v} = \lambda \vec{a} + \mu \vec{b}$.Substituting the vectors $\vec{a}$ and $\vec{b}$,we get $\vec{v} = \lambda(\hat{i} + \hat{j} + \hat{

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$3 \hat{i}-\hat{j}+3 \hat{k}$

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(C) Let $\vec{v} = \lambda \vec{a} + \mu \vec{b}$.Substituting the vectors $\vec{a}$ and $\vec{b}$,we get $\vec{v} = \lambda(\hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k}) = (\lambda + \mu)\hat{i} + (\lambda - \mu)\hat{j} + (\lambda + \mu)\hat{k}$.The projection of $\vec{v}$ on $\vec{c}$ is given by $\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.Given $\vec{c} = \hat{i} - \hat{j} - \hat{k}$,we have $|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.Thus,$\frac{(\lambda + \mu)(1) + (\lambda - \mu)(-1) + (\lambda + \mu)(-1)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.$\Rightarrow \lambda + \mu - \lambda + \mu - \lambda - \mu = 1$.$\Rightarrow \mu - \lambda = 1$,or $\mu = \lambda + 1$.Substituting $\mu$ back into $\vec{v}$,we get $\vec{v} = (2\lambda + 1)\hat{i} - \hat{j} + (2\lambda + 1)\hat{k}$.For $\lambda = 1$,we get $\vec{v} = 3\hat{i} - \hat{j} + 3\hat{k}$,which matches option $C$.

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,is given by

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