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# Angle between Two Vectors Formula

If the two vectors are given as $\vec{a}$ and $\vec{b}$ then its dot product is expressed as a.b. Suppose these two vectors are separated by angle $\theta$. To know what's the angle measurement we solve with the below formula The angle between two vectors formula is given by where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

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## Angle between two vectors Examples

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Lets see some examples on angle between two vectors:

### Solved Examples

Question 1: Find the angle between two vectors 3i + 4j - k and 2i - j + k.
Solution:

Given: $\vec{a}$ = 3i + 4j - k and $\vec{b}$ = 2i - j + k

The dot product is given by
a.b = (3i + 4j - k)(2i - j + k)
= (3)(2) + (4)(-1) + (-1)(1)
= 6-4-1
= 1

The magnitude of vectors is given by
|a| = $\sqrt{3^2 + 4^2 + (-1)^2}$ = $\sqrt{26}$ = 5.09
|b| = $\sqrt{2^2 + (-1)^2 + (1)^2}$ = $\sqrt{6}$ = 2.449

The angle between two vectors is

$\theta$ = cos-1 $\frac{a.b}{|a||b|}$

= cos-1 $\frac{1}{5.09 \times 2.449}$

= cos-1 $\frac{1}{12.465}$

= cos-1 0.0802
= 85.37o.

Question 2: Find the angle between two vectors 5i - j + k and i + j - k.
Solution:

Given: $\vec{a}$ = 5i - j + k and $\vec{b}$ = i + j - k
The dot product is given by
a.b = (5i - j + k)(i + j - k)
= (5)(1) + (-1)(1) + (1)(-1)
= 5-1-1
= 3
The magnitude of vectors is given by
|a| = $\sqrt{5^2 + (-1)^2 + (1)^2}$ = $\sqrt{26}$ = 5.09
|b| = $\sqrt{1^2 + (1)^2 + (-1)^2}$ = $\sqrt{3}$ = 1.73

The angle between two vectors is

$\theta$ = cos-1 $\frac{a.b}{|a||b|}$

= cos-1 $\frac{3}{5.09 \times 1.73}$

= cos-1 $\frac{3}{8.8057}$

= cos-1 0.372

= 68.16o.

*AP and SAT are registered trademarks of the College Board.