If bar{a}=hat{i}+2 hat{j}+hat{k},bar{b}=hat{i | vedclass

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(C) Let $\bar{r}$ be a vector in the plane of $\bar{a}$ and $\bar{b}$. Then,$\bar{r} = \bar{a} + m\bar{b}$ for some scalar $m$. $\bar{r} = (\hat{i} +

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$4 \hat{i}-\hat{j}+4 \hat{k}$

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(C) Let $\bar{r}$ be a vector in the plane of $\bar{a}$ and $\bar{b}$. Then,$\bar{r} = \bar{a} + m\bar{b}$ for some scalar $m$. $\bar{r} = (\hat{i} + 2\hat{j} + \hat{k}) + m(\hat{i} - \hat{j} + \hat{k}) = (1+m)\hat{i} + (2-m)\hat{j} + (1+m)\hat{k}$. The projection of $\bar{r}$ on $\bar{c}$ is given by $\frac{\bar{r} \cdot \bar{c}}{|\bar{c}|} = \frac{1}{\sqrt{3}}$. First,calculate $|\bar{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$. Now,$\bar{r} \cdot \bar{c} = (1+m)(1) + (2-m)(1) + (1+m)(-1) = 1+m + 2-m - 1-m = 2-m$. Thus,$\frac{2-m}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow 2-m = 1 \Rightarrow m = 1$. Substituting $m=1$ into the expression for $\bar{r}$: $\bar{r} = (1+1)\hat{i} + (2-1)\hat{j} + (1+1)\hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$. Checking the options,if we consider the projection as $\pm \frac{1}{\sqrt{3}}$,then $2-m = -1 \Rightarrow m = 3$. For $m=3$,$\bar{r} = (1+3)\hat{i} + (2-3)\hat{j} + (1+3)\hat{k} = 4\hat{i} - \hat{j} + 4\hat{k}$. This matches option $C$.

If $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$,$\bar{b}=\hat{i}-\hat{j}+\hat{k}$,and $\bar{c}=\hat{i}+\hat{j}-\hat{k}$,then a vector in the plane of $\bar{a}$ and $\bar{b}$,whose projection on $\bar{c}$ is $\frac{1}{\sqrt{3}}$,is

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