Question

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)`?

Answer 1

`theta = 0.37^("c")`

Explanation:

An angle between two vectors `ul(u)` and `ul(v)` can be found by the following formula

`theta =arccos((ul(u)*ul(v))/(abs(ulu) abs(ulv)))`

Where `ulu*ulv` is the scalar product of the vectors and `abs(ulu)` is the magnitude of `ulu`.

Let `ulu=((2),(2),(1))` and `ulv=((3),(4),(0))`

`ulu*ulv=2*3+2*4+1*0=14`

`absulu absulv = sqrt(2^2+2^2+1^1)sqrt(3^2+4^2) = 15`

`therefore theta = arccos(14/15)=0.37^("c")`

Answer 2

`21.0^o` to 1dp

Explanation:

The angle `theta` between two vectors `bb(vec A)` and `bb(vec B)` is related to the modulus (or magnitude) and scaler (or dot) product of `bb(vec A)` and `bb(vec B)` by the relationship:

`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)` and `bb(ul v)` be `theta` and let:

`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 `bb(vec A * vec B) = || bb(vec A) || \ || bb(vec B) || \ cos theta` we have:

`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`