# Angle between Two Vectors Formula

If the two vectors are assumed as

$$\begin{array}{l}\vec{a}\end{array}$$
and
$$\begin{array}{l}\vec{b}\end{array}$$
then the dot created is articulated as
$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
. Letâ€™s suppose these two vectors are separated by angleÂ Î¸.Â To know what’s the angleÂ measurement we solve with the below formula

we know that the dot product of two product is given as

$$\begin{array}{l}\vec{a}.\vec{b} =|\vec{a}||\vec{b}|cos\theta\end{array}$$

Thus, the angle between two vectors formula is given by

$$\begin{array}{l}\theta = cos^{-1}\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\end{array}$$

whereÂ Î¸ is the angle between

$$\begin{array}{l}\vec{a}\end{array}$$
and
$$\begin{array}{l}\vec{b}\end{array}$$

## Angle Between Two Vectors Examples

Letâ€™s see some samples on the angle between two vectors:

Example 1:

Â Compute the angle between two vectors 3i + 4j – k and 2i – j + k.

solution:

Let

$$\begin{array}{l}\vec{a}\end{array}$$
= 3i + 4j – k and

$$\begin{array}{l}\vec{b}\end{array}$$
= 2i – j + k

The dot product is defined as

$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
= (3i + 4j – k).(2i – j + k)

= (3)(2) + (4)(-1) + (-1)(1)

= 6-4-1

= 1

Thus,Â

$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
Â  = 1

The Magnitude of vectors is given by

$$\begin{array}{l}|\vec{a}| =\sqrt{(3^{2}+4^{2}+(-1)^{2})} =\sqrt{26}= 5.09\end{array}$$

$$\begin{array}{l}|\vec{b}| =\sqrt{(2^{2}+(-1)^{2}+1^{2})} =\sqrt{6}= 2.45\end{array}$$

The angle between the two vectors is

$$\begin{array}{l}\theta = cos^{-1}\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\end{array}$$

$$\begin{array}{l}\theta = cos^{-1}\frac{1}{(5.09)(2.45)}\end{array}$$

$$\begin{array}{l}\theta = cos^{-1}\frac{1}{12.47}\end{array}$$

$$\begin{array}{l}\theta = cos^{-1}(0.0802)\end{array}$$

$$\begin{array}{l}\theta = 85.39^{\circ}\end{array}$$

Example 2:Â

Find the angle between two vectors 5i – j + k and i + j – k.

Solution:

Let

$$\begin{array}{l}\vec{a}\end{array}$$
= 5i – j + k and

$$\begin{array}{l}\vec{b}\end{array}$$
= i + j â€“ k

The dot product is defined as

$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
= (5i – j + k)(i + j – k)

$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
= (5)(1) + (-1)(1) + (1)(-1)

$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
= 5-1-1

$$\begin{array}{l}\vec{a}. \vec{b}\end{array}$$
= 3

The Magnitude of vectors is given by

$$\begin{array}{l}|\vec{a}| =\sqrt{(5^{2}+(-1)^{2}+1^{2})} =\sqrt{27}= 5.19\end{array}$$

$$\begin{array}{l}|\vec{b}| =\sqrt{(1^{2}+1^{2}+(-1)^{2})} =\sqrt{3}= 1.73\end{array}$$

The angle between the two vectors is

$$\begin{array}{l}\theta = cos^{-1}\frac{\vec{a}.\vec{b}}{|\vec{a}||\vec{b}|}\end{array}$$

$$\begin{array}{l}\theta = cos^{-1}\frac{3}{(5.19)(1.73)}\end{array}$$

$$\begin{array}{l}\theta = cos^{-1}\frac{3}{8.97}\end{array}$$

$$\begin{array}{l}\theta = cos^{-1}(0.334)\end{array}$$

$$\begin{array}{l}\theta =70.48^{\circ}\end{array}$$