# Angle between Two Vectors Formula

If the two vectors are assumed as a⃗ and b⃗ then the dot created is articulated as a.b. Let’s suppose these two vectors are separated by angle θ. To know what’s the angle measurement we solve with the below formula

$a.b=\mid&space;a\mid&space;\mid&space;b\mid&space;cos\theta$

The angle between two vectors formula is given by
$\theta=cos^{-1}\frac{a.b}{\mid&space;a\mid&space;\mid&space;b\mid&space;}$

where θ is the angle between a⃗ a→ and b⃗ b→.

Examples

Let’s see some samples on angle between two vectors:

Problem 1: Compute the angle between two vectors 3i + 4j – k and 2i – j + k.
Answer:

a⃗ = 3i + 4j – k and

b⃗ = 2i – j + k

The dot product is articulated as
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

$\mid&space;a\mid=\sqrt{3^{2}+4^{2}+(-1^{2})}=\sqrt{26}=5.09$

$\mid&space;b\mid=\sqrt{2^{2}+(-1)^{2}+(1^{2})}=\sqrt{6}=2.449$

The angle between two vectors is

$\theta&space;=cos^{-1}\frac{a.b}{\mid&space;a\mid&space;\mid&space;b\mid&space;}$

$=cos^{-1}\frac{1}{5.09\times&space;2.449}$

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

$=cos^{-1}0.0802$

$=85.37^{\circ}$

Problem 2: Find the angle between two vectors 5i – j + k and i + j – k.
Answer:

Known:

a⃗ = 5i – j + k and

b⃗ = i + j – k

The dot product is articulated as
a.b = (5i – j + k)(i + j – k)

a.b = (5)(1) + (-1)(1) + (1)(-1)

a.b = 5-1-1

a.b = 3

The Magnitude of vectors is given by

$\mid&space;a\mid=\sqrt{5^{2}+(-1^{2})+1^{2}}=\sqrt{26}=5.09$

$\mid&space;b\mid=\sqrt{1^{2}+(1^{2})+(-1)^{2}}=\sqrt{3}=1.73$

The angle between two vectors is

$\theta&space;=cos^{-1}\frac{a.b}{\mid&space;a\mid&space;\mid&space;b\mid&space;}$

$=cos^{-1}\frac{3}{5.09\times&space;1.73}$

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

$=cos^{-1}0.372$

$=68.16^{\circ}$