Angle between Vectors?

Question

I could use help solving this question.

Find the angle between the vectors

u= 2i+5j and v=3i-2j-5k

The angle between the vectors is theta is approximately = fill in blank

(in radians)

Answer 1

cos Θ = u•v / ( |u| |v| )

u•v = 2(3)+5(-2)+0(-5) = -4

|u| = √(2²+5²) = √29

|v| = √(3²+2²+5²) = √38 .......... I ignored the negatives because they were getting squared.

cos Θ = -4/(√29 • √38) = -0.1205

Θ = 96.92°

remember cos 90° = 0

Answer 2

We're going to use the equation:

u dot v = |u||v|cos(theta)

Take the dot product:

2*3 + 5*-2 + 0*-5 = -4

Then find the magnitudes:

|u| = sqrt(2^2 + 5^2) = sqrt(29)

|v| = sqrt(3^2 + (-2)^2 + (-5)^2) = sqrt(38).

cos(theta) = -4 / ( sqrt(29) * sqrt(38) )

so take the inverse cosine on your calculator (I don't have one with me and don't know how to work this computer) to find theta.

Answer 3

Use the dot product.  Assuming i, j, and k are the basis units vectors i(1, 0, 0), j(0, 1, 0), k(0, 0, 1)

u(c, d).v(f, g) = cf + dg

u.v = |u| |v| cos a, where a is the angle between them.

|u| = sqrt(2^2 + 5^2 + 0) = sqrt(4 + 25) = sqrt(29)

|v| = sqrt(3^2 + (-2)^2 + (-5)^2) = sqrt(9+4+25) = sqrt(38)

u.v = 2 x 3 + 5 x -2 + 0 x -5 = 6 - 10 = -4

cos a = u.v / |u| |v|

cos a = -4 / sqrt(29)sqrt(38) = -4 / sqrt(1 102)

a ~= 1.6916 radians ~= 96.9 degrees

Answer 4

Given two vectors a and b, the way to find the angle between them is based on the following fact:

a dot b = |a|*|b|*cos(theta), where theta is the angle between the vectors a and b.

In this case, a = <2, 5, 0> and b = <3, -2, -5>.

a dot b = <2, 5, 0> dot <3, -2, -5> = 2*3 + 5*(-2) + 0*(-5) = -4

|a| = sqrt(4 + 25) = sqrt(29)

|b| = sqrt(9 + 4 + 25) = sqrt(38)

So, since a dot b = |a|*|b|*cos(theta),

-4 = sqrt(29)*sqrt(38)*cos(theta)

cos(theta) = -4/sqrt(1102)

theta = arccos(-4/sqrt(1102)) ~= 1.69 radians, or approximately 97 degrees.

Answer 5

u • v = | u | | v | cos Ө

6 - 10  - 25 = √29 √38 cos Ó¨

- 29 = √ 1102 cos Ө

cos Ó¨ = - 0.874

Ө = π ± 0.507