What is the angle between the given vector and the positive direction of the x-axis? i + √3j

Solution:

Given, the vector is i + √3j

We have to find the angle between the given vector and the positive x-axis.

The angle between two vectors is given by the formula

\(cos\theta=\frac{a_{1}b_{1}+a_{2}b_{2}}{(\sqrt{(a_{1})^{2}+(a_{2})^{2}})(\sqrt{(b_{1})^{2}+(b_{2})^{2}})}\)

The vector \((a_{1},a_{2})=\left \langle 1,\sqrt{3} \right \rangle\)

In two dimensional the x-axis vector form is \((b_{1},b_{2})=\left \langle 1,0 \right \rangle\)

So,\(cos\theta=\frac{(1\times1)+(\sqrt{3\times0})}{(\sqrt{(1)^{2}+(\sqrt{3})^{2}})(\sqrt{(1)^{2}+(0)^{2}})}\)

\(cos\theta=\frac{(1)+(0)}{(2)(1)}\)

\(cos\theta=\frac{1}{2}\)

Taking inverse,

\(\theta =cos^{-1}(\frac{1}{2})\)

\(\theta =cos^{-1}(cos\frac{\pi }{3})\)

\(\theta =\frac{\pi }{3}\)

Therefore, the angle is \(\theta =\frac{\pi }{3}\).

---

What is the angle between the given vector and the positive direction of the x-axis? i + √3j

Summary:

The angle between the given vector i + √3j and the positive direction of the x-axis is \(\theta =\frac{\pi }{3}\)