Question

A unit vector is perpendicular to both the vectors $\hat{i} +\hat{j}$ and $\hat{j} +\hat{k}$ is
A). $\dfrac{-\hat{i} -\hat{j} +\hat{k} }{\sqrt{3} }$
B). $\dfrac{\hat{i} +\hat{j} -\hat{k} }{3}$
C). $\dfrac{\hat{i} +\hat{j} +\hat{k} }{\sqrt{3} }$
D). $\dfrac{\hat{i} -\hat{j} +\hat{k} }{\sqrt{3} }$

Answer 1

Hint: In this question it is given that, we have to find the unit vector which is perpendicular to the vectors $\hat{i} +\hat{j}$ and $\hat{j} +\hat{k}$. So to find the solution we need to know that, if $\vec{a}$ and $\vec{b}$ be two vectors, then the cross product of two vectors ($\vec{a} \times \vec{b}$) is always perpendicular to the $\vec{a}$ and $\vec{b}$.
So by using the above concept we have to find the perpendicular vector.

Complete step-by-step solution:
Given vectors are $\hat{i} +\hat{j}$ and $\hat{j} +\hat{k}$, where $\hat{i} ,\hat{j} ,\hat{k}$ are the unit vectors.
Let us consider,
$\vec{a} =\hat{i} +\hat{j}$ and $\vec{b} =\hat{j} +\hat{k}$
Now we are going to find the cross product of $\vec{a} \ \text{and} \ \vec{b}$, for this let us consider the cross product of $\vec{a} \ \text{and} \ \vec{b}$ is $\vec{c}$.
Therefore,
$\vec{c} =\vec{a} \times \vec{b}$
$\Rightarrow \vec{c} =\begin{vmatrix}\hat{i} &\hat{j} &\hat{k} \\\ 1&1&0\\\ 0&1&1\end{vmatrix}$
Now by expanding the determinant, we get,
$\vec{c} =\hat{i} \left( 1\times 1-0\times 1\right) +\hat{j} \left( 0\times 0-1\times 1\right) +\hat{k} \left( 1\times 1-0\times 1\right)$
$\Rightarrow \vec{c} =\hat{i} -\hat{j} +\hat{k}$
Therefore $\vec{c}$ is the vector which is perpendicular to the vectors $\vec{a}$ and $\vec{b}$.
So in order to find the solution we need to find the unit vector along the direction of vector c.
So as we know that the unit vector of c is $\hat{c} =\dfrac{\vec{c} }{\left\vert \vec{c} \right\vert }$.
$\therefore \hat{c} =\dfrac{\hat{i} -\hat{j} +\hat{k} }{\left\vert \hat{i} -\hat{j} +\hat{k} \right\vert }$
$=\dfrac{\hat{i} -\hat{j} +\hat{k} }{\sqrt{1^{2}+\left( -1\right)^{2} +1^{2}} }$
$=\dfrac{\hat{i} -\hat{j} +\hat{k} }{\sqrt{1+1+1} }$
$=\dfrac{\hat{i} -\hat{j} +\hat{k} }{\sqrt{3} }$
Therefore, $\dfrac{\hat{i} -\hat{j} +\hat{k} }{\sqrt{3} }$ is the unit vector which is perpendicular to the vectors $\vec{a}$ and $\vec{b}$.
Hence the correct option is option D.

Note: To solve this type of question you need to know that if $\vec{a} =x_{1}\hat{i} +y_{1}\hat{j} +z_{1}\hat{k}$ and $\vec{b} =x_{2}\hat{i} +y_{2}\hat{j} +z_{2}\hat{k}$ be two vectors then their cross product is,
$\vec{a} \times \vec{b} =\begin{vmatrix}\hat{i} &\hat{j} &\hat{k} \\\ x_{1}&y;_{1}&z;_{1}\\\ x_{2}&y;_{2}&z;_{2}\end{vmatrix}$
And determinant of any vector is can be written as,
$\left\vert \vec{a} \right\vert =\sqrt{x^{2}_{1}+y^{2}_{1}+z^{2}_{1}}$.