# How do you use the definition of the scalar product, find the angles between the following pairs of vectors: A = - 3 i + j and B = 21 j - 3k?

##### 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=<<3, 1, 0>># and#vec v=<<0, 21, -3>>#

The modulus is given by;

# ||vec u|| = ||<<3, 1, 0>>|| = sqrt(3^2+1^2+0^2)=sqrt(9+1)=sqrt(10) #

# ||vec v|| = ||<<0, 21,-3>>|| = sqrt(0^2+21^2+(-3)^2)=sqrt(441+9)=sqrt(450) #

And the scaler product is:

# vec u * vec v = <<3, 1,0>> * <<0,21,-3>>#

# \ \ \ \ \ \ \ \ \ \ = (3)(0) + (1)(21) + (0)(-3)#

# \ \ \ \ \ \ \ \ \ \ = 21#

And so using

# 21 = sqrt(10) * sqrt(450) * cos theta #

# :. cos theta = (21)/sqrt(4500)#

# :. cos theta = (7sqrt(5))/50 #

# :. cos theta = 0.31304 ... #

# :. theta = 71.7568 ... °#

# :. theta = 71.8 °# (3sf)