# What is the angle between #<5,1,4 ># and #< 3,1,9 >#?

##### 1 Answer

#### Explanation:

The angle

# vec A * vec B = |A| |B| cos theta #

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem, let the angle betwen

#vec u=<<5,1,4>># and#vec v=<<3,1,9>>#

The modulus is given by;

# |vec u| = |<<5,1,4>>| = sqrt(5^2+1^2+4^2)=sqrt(25+1+16)=sqrt(42) #

# |vec v| = |<<3,1,9>>| = sqrt(3^2+1^2+9^2)=sqrt(9+1+81)=sqrt(91) #

And the scaler product is:

# vec u * vec v = <<5,1,4>> * <<3,1,9>>#

# \ \ \ \ \ \ \ \ \ \ = (5)(3) + (1)(1) +(4)(9) #

# \ \ \ \ \ \ \ \ \ \ = 15+1+36#

# \ \ \ \ \ \ \ \ \ \ = 52#

And so using

# 52 = sqrt(42) * sqrt(91) * cos theta #

# :. cos theta = (52)/(sqrt(42) * sqrt(91))#

# :. cos theta = 0.841120 ... #

# :. theta = 32.741412 °#

# :. theta = 32.7 °# (3sf)