Question

A unit vector perpendicular to both the vectors $i+j$ and $j+k$ is

• A. $\frac{\left(-i-j+k\right)}{\sqrt{3}}$
• B. $\left(\frac{i+j-k}{3}\right)$
• C. $\frac{\left(i+j+k\right)}{\sqrt{3}}$
• D. $\frac{\left(i-j+k\right)}{\sqrt{3}}$

Answer 1

The correct option is D

$\frac{\left(i-j+k\right)}{\sqrt{3}}$

Explanation for correct answer:

Step-1: Express the given data.

Let, the given $\overrightarrow{a}=i+j\&\overrightarrow{b}=j+k$

Let $\overrightarrow{c}$ is a unit vector which is perpendicular to the vector $\overrightarrow{a}\&\overrightarrow{b}$.

Step-2: Finding the unit vector $\overrightarrow{c}$.

Since, vector $\overrightarrow{c}$ is perpendicular to the vector $\overrightarrow{a}\&\overrightarrow{b}$

$\Rightarrow$$\overrightarrow{c}=\overrightarrow{a}\times \overrightarrow{b}$

$\Rightarrow$$\overrightarrow{c}=\left|\begin{array}{ccc}i& j& k\\ 1& 1& 0\\ 0& 1& 1\end{array}\right|$

$\Rightarrow$$\overrightarrow{c}=i-j+k$

Therefore, unit vector along the vector $\overrightarrow{c}$ is $\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}$

$\Rightarrow$$\hat{c}=\frac{i-j+k}{\sqrt{3}}$

Hence, option $\left(D\right)$ is correct.