$\vec{a}=2 \hat{i}-\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-\hat{k}, \quad \vec{c}=\hat{i}+\hat{j}-2 \hat{k}$. A vector coplanar with $\vec{b}$ and $\vec{c}$, whose projection on $\vec{a}$ is of magnitude $\sqrt{\frac{2}{3}}$ is:
Answer & explanation
Correct answer: option 1
Let the required vector be $\vec{r}$
Then, $\vec{r}=x_1 \vec{b}+x_2 \vec{c}$ and $\vec{r} . \hat{a}= \pm \sqrt{\frac{2}{3}}$
$\Rightarrow \vec{r} . \hat{a}= \pm \sqrt{\frac{2}{3}}|\vec{a}|=2$
Now, $\vec{r} . \vec{a}=x_1 \vec{d} . \vec{b}+x_2 \vec{a} . \vec{c}$
$\Rightarrow \pm 2=x_1(2-2-1)+x_2(2-1-2)$
$\Rightarrow x_1+x_2$ = -2 or 2
If $x_1+x_2=-2$, then
$\vec{r}=x_1(\hat{i}+2 \hat{j}-\hat{k})+x_2(\hat{i}+\hat{j}-2 \hat{k})$
$=\hat{i}\left(x_1+x_2\right)+\hat{j}\left(2 x_1+x_2\right)-\hat{k}\left(x_1+2 x_2\right)$
$=-2 \hat{i}+\hat{j}\left(x_1-2\right)-\hat{k}\left(x_2-2\right)$
$=-2 \hat{i}+\hat{j}\left(x_1-2\right)-\hat{k}\left(-4-x_1\right)$
where $x_1 \in R$
If $x_1+x_2=2$, then
$\vec{r}=x_1(\hat{i}+2 \hat{j}-\hat{k})+x_2(\hat{i}+\hat{j}-2 \hat{k})$
$=\hat{i}\left(x_1+x_2\right)+\hat{j}\left(2 x_1+x_2\right)-\hat{k}\left(x_1+2 x_2\right)$
$=2 \hat{i}+\hat{j}\left(x_1+2\right)-\hat{k}\left(4-x_1\right)$
Hence (1) is correct answer.