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,
$$\left\vert \vec{a} \right\vert =\sqrt{x^{2}_{1}+y^{2}_{1}+z^{2}_{1}}$$.