Mathematics - Vectors Question with Solution | TestHub

Given $\bar{a}=\bar{i}+\bar{j}-\bar{k}$ and $\bar{b}=\bar{i}-\bar{j}+\bar{k}$.

Since $\bar{c}$ is coplanar with $\bar{a}$ and $\bar{b}$, $\bar{c}=x\bar{a}+y\bar{b}$.

Also, $\bar{c}\perp \bar{a}$, so $\bar{c}\cdot \bar{a}=0$.

$(x\bar{a}+y\bar{b})\cdot \bar{a}=0\Rightarrow x|\bar{a}{|}^2+y(\bar{a}\cdot \bar{b})=0$.

$|\bar{a}{|}^2=1^2+1^2+(-1{)}^2=3$.

$\bar{a}\cdot \bar{b}=(1)(1)+(1)(-1)+(-1)(1)=1-1-1=-1$.

So, $3x-y=0\Rightarrow y=3x$.

$\bar{c}=x(\bar{i}+\bar{j}-\bar{k})+3x(\bar{i}-\bar{j}+\bar{k})=x(\bar{i}+\bar{j}-\bar{k}+3\bar{i}-3\bar{j}+3\bar{k})=x(4\bar{i}-2\bar{j}+2\bar{k})$.

Since $\bar{c}$ is a unit vector, $|\bar{c}|=1$.

$|x(4\bar{i}-2\bar{j}+2\bar{k})|=1\Rightarrow |x|\sqrt{4^2+(-2{)}^2+2^2}=1\Rightarrow |x|\sqrt{16+4+4}=1\Rightarrow |x|\sqrt{24}=1$.

Let $x=\frac{1}{\sqrt{24}}=\frac{1}{2\sqrt{6}}$.

$\bar{c}=\frac{1}{2\sqrt{6}}(4\bar{i}-2\bar{j}+2\bar{k})=\frac{1}{\sqrt{6}}(2\bar{i}-\bar{j}+\bar{k})$.

We need a unit vector $\bar{d}$ perpendicular to both $\bar{a}$ and $\bar{c}$.

Thus, $\bar{d}$ is parallel to $\bar{a}\times \bar{c}$.

$\bar{a}\times \bar{c}=(\bar{i}+\bar{j}-\bar{k})\times \frac{1}{\sqrt{6}}(2\bar{i}-\bar{j}+\bar{k})$

$$=\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\bar{i}& \bar{j}& \bar{k}\\ 1& 1& -1\\ 2& -1& 1\end{array}\right|$$

$$=\frac{1}{\sqrt{6}}[\bar{i}(1-1)-\bar{j}(1-(-2))+\bar{k}(-1-2)]$$

$$=\frac{1}{\sqrt{6}}[0\bar{i}-3\bar{j}-3\bar{k}]=\frac{-3}{\sqrt{6}}(\bar{j}+\bar{k})$$

The unit vector $\bar{d}$ is $\frac{\bar{a}\times \bar{c}}{|\bar{a}\times \bar{c}|}$.

$|\frac{-3}{\sqrt{6}}(\bar{j}+\bar{k})|=\frac{3}{\sqrt{6}}\sqrt{1^2+1^2}=\frac{3}{\sqrt{6}}\sqrt{2}=\frac{3}{\sqrt{3}}=\sqrt{3}$.

So, $\bar{d}=\frac{\frac{-3}{\sqrt{6}}(\bar{j}+\bar{k})}{\sqrt{3}}=\frac{-3}{\sqrt{18}}(\bar{j}+\bar{k})=\frac{-3}{3\sqrt{2}}(\bar{j}+\bar{k})=\frac{-1}{\sqrt{2}}(\bar{j}+\bar{k})$.

The direction can be positive or negative, so $\bar{d}=\pm \frac{1}{\sqrt{2}}(\bar{j}+\bar{k})$.

Comparing with the options, option C is $\frac{1}{\sqrt{2}}(\bar{j}+\bar{k})$.

The final answer is $\boxed{\frac{1}{\sqrt{2}}(\bar{j}+\bar{k})}$.