What is the angle θab between a⃗ and b⃗ ? Express your answer using one significant figure.

Solution:

In order to determine the angle between two vectors we have to draw them with their beginning from a common point. The angle between these two vectors is the smallest of two possible angles.

Using the definition of dot product

\(\overrightarrow{a}.\overrightarrow{b}=|\overrightarrow{a}|.|\overrightarrow{b}|.cos\Theta _{\overrightarrow{a},\overrightarrow{b}}\)

It can be written as

\(cos\Theta _{\overrightarrow{a},\overrightarrow{b}}=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|.|\overrightarrow{b}|}\)

If \(\overrightarrow{a}=(a_{1},a_{2},a_{3})\: and\: \overrightarrow{b}=(b_{1},b_{2},b_{3})\)

\(cos\Theta _{\overrightarrow{a},\overrightarrow{b}}=\frac{a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}{\sqrt{a_{2}^{2}+a_{1}^{2}+a_{3}^{2}}.\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\)

Therefore, the angle is \(cos\Theta _{\overrightarrow{a},\overrightarrow{b}}=\frac{a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}{\sqrt{a_{2}^{2}+a_{1}^{2}+a_{3}^{2}}.\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\).

What is the angle θab between a⃗ and b⃗ ? Express your answer using one significant figure.

Summary:

The angle θab between a and b is \(cos\Theta _{\overrightarrow{a},\overrightarrow{b}}=\frac{a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}}{\sqrt{a_{2}^{2}+a_{1}^{2}+a_{3}^{2}}.\sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\).