Question

Which of the following is not a unit vector?
A. $\dfrac{{2\hat i + 3\hat j + 2\hat k}}{{\sqrt {17} }}$
B. $\dfrac{{2\hat i - 3\hat j - \hat k}}{{\sqrt {17} }}$
C. $\dfrac{{\hat i - \hat j - \hat k}}{{\sqrt 3 }}$
D. $\dfrac{{\hat i + \hat k}}{{\sqrt 3 }}$

Answer 1

Hint: We know that a unit vector is a vector whose magnitude is equal to $1$ . Here, we will find the magnitude of each vector from the given option and whose magnitude will be unity that will be a unit vector.

Complete step by step answer:
For a given vector in the form of $a\hat i + b\hat j + c\hat k$ the magnitude of such vectors is given by$\sqrt {{a^2} + {b^2} + {c^2}} $. $\hat i$, $\hat j$ And $\hat k$ are the unit vectors of length one in the direction of $X$ axis, $Y$ axis and $Z$ axis. And $a$ $b$ $c$ are the magnitudes of a vector which makes intercept on these axes respectively.
Firstly, we will find the magnitude of given vector $\dfrac{{2\hat i + 3\hat j + 2\hat k}}{{\sqrt {17} }}$
Magnitude $ = \dfrac{{\sqrt {4 + 9 + 4} }}{{\sqrt {17} }}= 1$ Unit.
Hence a given vector in option A is a unit vector.
Now, magnitude of vector $\dfrac{{2\hat i - 3\hat j - \hat k}}{{\sqrt {17} }}$ is given by
$\Rightarrow \dfrac{{\sqrt {1 + 1 + 1} }}{{\sqrt 3 }}= 1$ Unit.
Hence a given vector in option B is a unit vector.
Now, magnitude of vector $\dfrac{{\hat i - \hat j - \hat k}}{{\sqrt 3 }}$ is given by
$\Rightarrow \dfrac{{\sqrt {1 + 1 + 1} }}{{\sqrt 3 }}=1$ Unit.
Hence a given vector in option C is also a unit vector.
Now, magnitude of vector $\dfrac{{\hat i + \hat k}}{{\sqrt 3 }}$ is given by
$\Rightarrow \dfrac{{\sqrt {1 + 1} }}{{\sqrt 3 }}=\sqrt {\dfrac{2}{3}} $
Which does not equal unity. So, the vector given in option D is not a unit vector.

Hence, option D is the answer.

Note: The only purpose of a unit vector is to give the idea of direction in which a vector is headed. A unit vector is also known as a Direction vector. There is a radial unit vector which lies along the surface and there is a normal unit vector which is perpendicular to the surface.