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$\vec{a}=\hat{i}+\hat{j}+\hat{k}$, and $\vec{b}=\hat{i}-\hat{j}+\widehat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ are three vectors then vector $\overline{\mathrm{r}}$ in the plane of $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, whose projection on $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$, is given by
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Verified Answer
The correct answer is:
$(2 \mathrm{t}+1) \hat{\mathrm{i}}-\hat{\mathrm{j}}+(2 \mathrm{t}+1) \widehat{\mathrm{k}}, \forall \mathrm{t} \in \mathrm{R}$
$\vec{r}=t \vec{a}+u \vec{b}$ [as vector lies in the plane of $\vec{a}$ and $\vec{b}$ where $t$ and $u$ are scalars]
$\begin{aligned} & \Rightarrow \vec{r}=t(\hat{i}+\hat{j}+\hat{k})+u(\hat{i}-\hat{j}+\hat{k}) \\ & =(t+u) \hat{i}+(t-u) \hat{j}+(t+u) \hat{k} \ldots(i) \\ & \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \Rightarrow \frac{(t+u)-(t-u)-(t+u)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow-(t-u)=1 \Rightarrow u=t+1 \ldots \ldots . .(i i)\end{aligned}$
from (i) and (ii)
$\vec{r}=(2 t+1) \hat{i}-\hat{j}+(2 t+1) \widehat{k}$
$\begin{aligned} & \Rightarrow \vec{r}=t(\hat{i}+\hat{j}+\hat{k})+u(\hat{i}-\hat{j}+\hat{k}) \\ & =(t+u) \hat{i}+(t-u) \hat{j}+(t+u) \hat{k} \ldots(i) \\ & \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \Rightarrow \frac{(t+u)-(t-u)-(t+u)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\ & \Rightarrow-(t-u)=1 \Rightarrow u=t+1 \ldots \ldots . .(i i)\end{aligned}$
from (i) and (ii)
$\vec{r}=(2 t+1) \hat{i}-\hat{j}+(2 t+1) \widehat{k}$
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