Question

# The unit vector in the direction of $\widehat i + \widehat j + \widehat k$ is be?a)$\dfrac{1}{{\sqrt 3 }}(\widehat i + \widehat j + \widehat {k)}$b)$\sqrt 3 (\widehat i + \widehat j + \widehat {k)}$c)$\dfrac{1}{{\sqrt 2 }}(\widehat i + \widehat j + \widehat {k)}$d)$\sqrt 2 (\widehat i + \widehat j + \widehat {k)}$

Hint: Here, we will use the formulae of unit vector in the direction of unit vector to solve.

Given,
Vector is$\widehat i + \widehat j + \widehat k$. Now, we need to find the unit vector in the direction of $\widehat i + \widehat j + \widehat k$. As, we know that a unit vector is given by dividing the vector by its magnitude, so the resulting vector has magnitude 1and is in same direction as the original vector.
Let$\widehat a = \widehat i + \widehat j + \widehat k$, Now magnitude of $\widehat a$is
$\begin{gathered} \Rightarrow \left| {\widehat a} \right| = \sqrt {{1^2} + {1^2} + {1^2}} \\ \Rightarrow \left| {\widehat a} \right| = \sqrt 3 \\ \end{gathered}$
As, we know the unit vector in the direction of $\widehat a$ is $\dfrac{{\widehat a}}{{\left| {\widehat a} \right|}}$.Now, the unit vector in the direction of $\widehat i + \widehat j + \widehat k$ will be
$\Rightarrow \dfrac{{\widehat i + \widehat j + \widehat k}}{{\sqrt 3 }}$
Therefore, the unit vector in the direction of $\widehat i + \widehat j + \widehat k$ is $\dfrac{{\widehat i + \widehat j + \widehat k}}{{\sqrt 3 }}$.
Hence, the required option for the given question is ‘A’.

Note: Here, we have been asked to find the unit vector in the direction of the vector, hence the sign of the unit vector is positive elsewhere. If we need to find the unit vector in the opposite direction of the given vector then the sign of the unit vector will be negative.