Question

The unit vector perpendicular to the plane containing the vectors $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}-\hat{k}$ is

• A. $\pm \frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
• B. $\pm \frac{1}{\sqrt{2}}(\hat{i}-\hat{j})$
• C. $\pm \frac{1}{\sqrt{2}}(\hat{j}-\hat{k})$
• D. $\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$

Answer 1

Answer: (D)

$\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$

We know that the cross product of any two vectors yields a vector which is perpendicular to both vectors. Therefore, for two vectors $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}-\hat{k}$ if $\overrightarrow{C}$ is the vector perpendicular to both, then we have:

$\overrightarrow{C}=\left[\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 1& -1& -1\end{array}\right]=\hat{i}\left[\right(1\times -1)-(-1\times 1\left)\right]-\hat{j}\left[\right(1\times -1)-(1\times 1\left)\right]+\hat{k}\left[\right(1\times -1)-(1\times 1\left)\right]=\hat{i}[-1-(-1\left)\right]-\hat{j}[-1-1]+\hat{k}[-1-1]=\hat{i}[-1+1]-\hat{j}(-2)+\hat{k}(-2)=-2\hat{j}-2\hat{k}$

Now, unit vector in the direction of $\overrightarrow{C}$ is $\frac{\overrightarrow{C}}{\left|\overrightarrow{C}\right|}$, therefore,

$\left|\overrightarrow{C}\right|=\sqrt{0^2+(-2{)}^2+(-2{)}^2}=\sqrt{4+4}=\sqrt{8}=\pm 2\sqrt{2}$

Thus, the required unit vector is:

$\frac{\overrightarrow{C}}{\left|\overrightarrow{C}\right|}=\frac{-2\hat{j}-2\hat{k}}{2\sqrt{2}}=\pm \frac{2(\hat{j}+\hat{k})}{2\sqrt{2}}=\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$

Hence, the unit vector is $\pm \frac{1}{\sqrt{2}}(\hat{j}+\hat{k})$.