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

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Aicharya · Accepted Answer

Let $\overline{\mathrm{r}}$ be the vector coplanar to $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$. Then, $\begin{aligned} \overline{\mathrm{r}} & =\overline{\mathrm{a}}+\mathrm{m} \overline{\mathrm{b}} \ & =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}})+\mathrm{m}(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \ & =\hat{\mathrm{i}}(1+\mathrm{m})+\hat{\mathrm{j}}(2-\mathrm{m})+\hat{\mathrm{k}}(1+\mathrm{m}) \end{aligned}$ Since the projection of $\overline{\mathrm{r}}$ along $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$, $\begin{aligned} & \frac{\overline{\mathrm{r}} \cdot \overline{\mathrm{c}}}{\overline{\mathrm{c}}}= \pm \frac{1}{\sqrt{3}} \ & \Rightarrow \frac{(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})}{\sqrt{3}}= \pm \frac{1}{\sqrt{3}} \ & \Rightarrow(1+\mathrm{m})+(2-\mathrm{m})-(1+\mathrm{m})= \pm 1 \ 	herefore \quad & \mathrm{m}=3 	ext { or } \mathrm{m}=1 \end{aligned}$ Substituting $\mathrm{m}=3$ in equation (i), we get $\begin{aligned} & \overline{\mathrm{r}}=\hat{\mathrm{i}}(1+3)+\hat{\mathrm{j}}(2-3)+\hat{\mathrm{k}}(1+3) \ & \Rightarrow \overline{\mathrm{r}}=4 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \end{aligned}$

If $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}, \bar{b}=\hat{i}-\hat{j}+\hat{k}, \bar{c}=\hat{i}+\hat{j}-\hat{k}$,…

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