Question

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{c}=\hat{i}+\hat{j}-\hat{k}$. A vector in the plane of $\overrightarrow{a}$ and $\overrightarrow{b}$ whose projection on $\overrightarrow{c}$ is $-\frac{1}{\sqrt{2}}$, is

Answer 1

Let vector be $\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k}$

Now as $\overrightarrow{d}$ lies in plane of $\overrightarrow{a}$ and $\overrightarrow{b}$

$\therefore \overrightarrow{b}.(\overrightarrow{a}\times \overrightarrow{d})=0$

$\Rightarrow \left|\begin{array}{ccc}1& -1& 1\\ 1& 2& 1\\ x& y& z\end{array}\right|=0$

$C1\to C1+C2,C2\to C2+C3$

$\left|\begin{array}{ccc}0& 0& 1\\ 3& 3& 1\\ x+y& y+z& z\end{array}\right|=0$

$3(y+z)-3(x+y)=0\Rightarrow x=z$

Now, projection of $\overrightarrow{d}$ on $\overrightarrow{c}$ is $\frac{\overrightarrow{d}.\overrightarrow{c}}{|\overrightarrow{c}|}\Rightarrow \frac{(x\hat{i}+y\hat{j}+x\hat{k})(\hat{i}+\hat{j}-\hat{k})}{\sqrt{1^2+1^2+(-1{)}^2}}=-\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{y}{\sqrt{3}}=-\frac{1}{\sqrt{2}}\Rightarrow y=-\sqrt{\frac{3}{2}}$

$\therefore \overrightarrow{d}=\lambda \hat{i}-\sqrt{\frac{3}{2}}\hat{j}+\lambda \hat{k}....($where $x=z=\lambda$ and $\lambda ϵR)$