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

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(B) Since $\vec{r}$ lies in the plane of $\vec{a}$ and $\vec{b}$,we can write $\vec{r} = t\vec{a} + u\vec{b}$ for some scalars $t$ and $u$.$\vec{r} =

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$(2t+1)\hat{i}-\hat{j}+(2t+1)\hat{k}, \forall t \in R$

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(B) Since $\vec{r}$ lies in the plane of $\vec{a}$ and $\vec{b}$,we can write $\vec{r} = t\vec{a} + u\vec{b}$ for some scalars $t$ and $u$.$\vec{r} = t(\hat{i}+\hat{j}+\hat{k}) + u(\hat{i}-\hat{j}+\hat{k}) = (t+u)\hat{i} + (t-u)\hat{j} + (t+u)\hat{k} \dots (i)$Given that the projection of $\vec{r}$ on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,we have $\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}}$.$|\vec{c}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$.$\vec{r} \cdot \vec{c} = (t+u)(1) + (t-u)(-1) + (t+u)(-1) = t+u - t+u - t-u = u-t$.So,$\frac{u-t}{\sqrt{3}} = \frac{1}{\sqrt{3}} \Rightarrow u-t = 1 \Rightarrow u = t+1$.Substituting $u = t+1$ into equation $(i)$:$\vec{r} = (t + t + 1)\hat{i} + (t - (t + 1))\hat{j} + (t + t + 1)\hat{k}$$\vec{r} = (2t+1)\hat{i} - \hat{j} + (2t+1)\hat{k}$.

If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$,and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ are three vectors,then the vector $\vec{r}$ 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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