Question

Two vectors, A and B, have magnitudes of 10 and 15. The cross product of these vectors has magnitude 32. What is the angle between vectors A and B?

Answer 1

Answer:

The Angle between the vector A and B is 12.32°

Explanation:

Given data

|A×B|=32

|A|=10

|B|=15

To find

Angle α

Solution

From Cross product properties we know that

$|A*B|=|A||B|Sin\alpha \\32=(10)(15)Sin\alpha\\Sin\alpha=32/150\\\alpha =Sin^{-1}(32/150)\\\alpha =12.32^{o}$

The Angle between the vector A and B is 12.32°

Answer 2

Answer:

12.32°

Explanation:

The vector product of two vectors A and B (written as A x B) is the vector C that is perpendicular to the two vectors. The magnitude of such vector C (written as |C| ) is given by the product of the magnitudes of the two vectors, A and B (written as |A| |B|), and the sine of the angle θ between them. This can be represented as follows;

|C| = |A x B| = |A||B| sin θ    --------------------(i)

From the question;

The magnitude of the cross product, |A x B| = 32

The magnitude of vector A, |A| = 10

The magnitude of vector B, |B| = 15

Substitute these values into equation (i) as follows;

32 = 10 x 15 sin θ

32 = 150 sin θ

sin θ = 32 / 150

sin θ = 0.2133

Solve for θ;

θ = sin⁻¹ (0.2133)

θ = 12.32°

Therefore, the angle between these vectors is 12.32°