# What is the angle between the vectors #2bb(ul hat i)+2bb(ul hat j)+bb(ul hat k)# and #3bb(ul hat i)+4bb( ul hat j)#?

##### 2 Answers

#### Explanation:

An angle between two vectors

Where

Let

#### Explanation:

The angle

# bb(vec A * vec B) = || bb(vec A) || \ || bb(vec 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 between

#bb(ul u) = 2bb(ul hat i)+2bb(ul hat j)+bb(ul hat k)#

#bb(ul v) = 3bb(ul hat i)+4bb( ul hat j)#

The vector norm is given by;

# || bb(ul u) || = || 2bb(ul hat i)+2bb(ul hat j)+bb(ul hat k) || =sqrt(2^2+2^2+1^2) = 3#

# || bb(ul v) || = || 3bb(ul hat i)+4bb( ul hat j) || \ \ \ \ \ \ \ \ = sqrt( 3^2+4^2) \ \ \ \ \ \ \ \ = 5#

And the scaler (or "dot") product is:

# bb(ul u * ul v) = (2bb(ul hat i)+2bb(ul hat j)+bb(ul hat k)) * (3bb(ul hat i)+4bb( ul hat j))#

# \ \ \ \ \ \ \ \ \ \ \ = (2)(3) + (2)(4) + (1)(0) #

# \ \ \ \ \ \ \ \ \ \ \ = 6 + 8 + 0 #

# \ \ \ \ \ \ \ \ \ \ \ = 14 #

And so using

# 14 = 3 * 5* cos theta #

# :. cos theta = 14/15 #

# => theta = 21.03946 ... ^o #

So the acute angle between the vectors is:

# theta_("acute") = 21.03946 ...^o #