I am confused! Vectors!?

Question

What are the angles between these vectors?

1. v=4i+2j, w=5j+k
2. v=3i-4j-k, w=-2j

I got 0 degrees for both, did I do something wrong?
Please show your work thanks!

Answer 1

The best way to get the angle between vectors is to use the dot product. Recall that
v dot w=length of v*length of w*cosine(angle between them).

You can easily find the angle with this formula.

For problem 1,

v dot w= 4*0+2*5+0*1=10

The length of v=sqrt(4^2+2^2)=sqrt(20)

The length of w=sqrt(5^2+1^2)=sqrt(26)

So our formula gives us:

10=sqrt(20)*sqrt(26)*cos(angle)

After calculation, the angle=64.0 degrees

Same thing for problem 2

v dot w=8
length v=sqrt(26)
length w=2
(Do you see how I got these numbers?)

8=2sqrt(26)*cos(angle)
angle=38.3 degrees.

Answer 2

I'll guess that you know about the dot product between two vectors. We can write the dot product in terms of the angle between the two vectors:

v•w = |v|*|w|*cos θ

By |v| I mean the length of the vector, determined by the Pythagorean theorem. Thus if v = ai + bj + ck, then |v| = √(a² + b² + c²). θ is the angle between the two vectors. This means we can rearrange this as:

cos θ = (v•w)/(|v|*|w|)

So once we compute the right-hand side, we can take the arccosine. That'll give us our angle.

1. v•w = (4i + 2j + 0k)•(0i + 5j + 1k) = (4*0) + (2*5) + (0*1) = 10
|v| = √(4² + 2² + 0²) = √20 = 2√5
|w| = √(0² + 5² + 1²) = √26
Thus cos θ = (10)/[(2√5)*(√26)] = 5/√130 ≈ 0.4385.
Hence θ ≈ arccos(0.4385) ≈ 63.99° ≈ 1.12 radians.

2. v•w = (3i - 4j - 1k)•(0i - 2j + 0k) = (3*0) + (-4 * -2) + (-1*0) = 8
|v| = √(3² + (-4)² + (-1)²) = √26
|w| = √(0² + (-2)² + 0²) = 2
Thus cos θ = (8)/[(√26)*(2)] = 4/√26 ≈ 0.7845.
Hence θ ≈ arccos(0.7845) ≈ 38.33° ≈ 0.67 radians.

Answer 3

Question 1
let v and w be vectors.
v = (4 2 0) and w = (0 5 1)
v . w = 10
v . w = |v| |w| cos Ø
10 = √20 x √26 cos Ø
cos Ø = 10 / (√20 x √26)
cos Ø = 0 . 438
Ø = 64°
Question 2
v = (3 - 4 - 1) and w = (0 - 2 0)
v.w = 8
|v| = √26 and |w| = 2
8 = √26 x 2 cos Ø
cos Ø = 4 / √26
Ø = 38.3°