The product of the magnitudes of two vectors A & B is 12. Scalar product of A & B is 6√3. Find the angle between A & B. When the scalar product will be zero

The scalar product of two vectors A and B is given by the formula:

A.B = |A||B|cosθ

where:

• A.B is the scalar product of A and B,
• |A| and |B| are the magnitudes of vectors A and B respectively,
• θ is the angle between A and B.

Given that the product of the magnitudes of A and B is 12, and the scalar product of A and B is 6√3, we can substitute these values into the formula:

6√3 = 12cosθ

Solving for cosθ gives:

cosθ = 6√3 / 12 = √3 / 2

The angle θ whose cosine is √3 / 2 is 30 degrees. So, the angle between vectors A and B is 30 degrees.

The scalar product of two vectors is zero when the vectors are perpendicular to each other. This is because the cosine of 90 degrees is zero, so if θ = 90 degrees, then A.B = 0. Therefore, the scalar product will be zero when the angle between vectors A and B is 90 degrees.