Question

Write down a unit vector in XY-plane, making an angle of ${30}^{\circ }$ with the
positive direction of x-axis.

Answer 1

Hint: If a unit vector is making an angle $\theta$ with the positive direction of x-axis, then it’s component along the x-axis is $\cos \theta$ and it’s component along the positive y-axis is $\sin \theta$. This unit vector will be written as $\cos \theta \hat{i}+\sin \theta \hat{j}$.

Before proceeding with the question, we must know the formula that will be required to solve this question. In vectors, if a unit vector is making an angle $\theta$ with the positive x-axis, then the vector can be written as $\cos \theta \hat{i}+\sin \theta \hat{j}$. The x component of this vector is $\cos \theta$ and the y component of this vector is $\sin \theta$.
In the question, we have to write down the unit vector making an angle ${30}^{\circ }$ with the positive x-axis.

The x component of this vector will be $\cos {30}^{\circ }$ and the y component of this vector will be $\sin {30}^{\circ }$. So, this unit vector which is making an angle ${30}^{\circ }$ with the positive x-axis can be written as $$\cos {30}^{\circ }\hat{i}+\sin {30}^{\circ }\hat{j}$$.
From trigonometry, we have $\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$ and $\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$. Substituting these values in the above vector, we get,
${\displaystyle \frac{\sqrt{3}}{2}}\hat{i}+{\displaystyle \frac{1}{2}}\hat{j}$
There is one more possible vector that can make an angle of ${30}^{\circ }$ with the positive x-axis.
That vector is as shown below,

The x component of this vector will be $\cos {30}^{\circ }$ and the y component of this vector will be $\sin {30}^{\circ }$. But in this vector, the y component will be along the negative y-axis. So, this unit vector which is making an angle ${30}^{\circ }$ with the positive x-axis can be written as
$$\cos {30}^{\circ }\hat{i}-\sin {30}^{\circ }\hat{j}$$.
From trigonometry, we have $\cos {30}^{\circ }={\displaystyle \frac{\sqrt{3}}{2}}$ and $\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$. Substituting these values in the above vector, we get,
${\displaystyle \frac{\sqrt{3}}{2}}\hat{i}-{\displaystyle \frac{1}{2}}\hat{j}$
So, the two possible answers are ${\displaystyle \frac{\sqrt{3}}{2}}\hat{i}+{\displaystyle \frac{1}{2}}\hat{j}$ and ${\displaystyle \frac{\sqrt{3}}{2}}\hat{i}-{\displaystyle \frac{1}{2}}\hat{j}$.

Note: In this question, we were given a unit vector which is making an angle of ${30}^{\circ }$ with the positive x-axis. We wrote this vector by using the formula $\cos \theta \hat{i}+\sin \theta \hat{j}$. But in case we are given a vector which is also having a magnitude, let us say $r$and making an angle $\theta$ with the positive x-axis, we will write that vector by using the formula
$r(\cos \theta \hat{i}+\sin \theta \hat{j})$.