vector a has magnitude 1 and vector b has magnitude 16m. Their scalar product is 90.0m^2. What is the magnitude of the vector product between these 2 vectors?

scalar product or dot product:

a∙b = |a| |b|cosØ, where Ø is the angle between then
1(16)cosØ = 90
cosØ= 90/16 = 5.625 , which is not possible, since cos(anything) falls between -1 and +1

To find the magnitude of the vector product between two vectors, we need to use the formula:

|A x B| = |A| * |B| * sin(theta)

Where |A x B| is the magnitude of the vector product of vectors A and B, |A| and |B| are the magnitudes of vectors A and B respectively, and theta is the angle between A and B.

In this case, we are given the magnitudes of vectors A and B, but not the angle theta. However, we are given the scalar product of vectors A and B, which can help us find the angle between them.

The scalar product (also known as the dot product) of two vectors A and B is given by the formula:

A · B = |A| * |B| * cos(theta)

Given in the problem that the scalar product of vectors A and B is 90.0, we can rewrite the formula as:

90.0 = 1 * 16 * cos(theta)

Simplifying the equation, we have:

cos(theta) = 90.0 / (1 * 16)
cos(theta) = 90.0 / 16
cos(theta) = 5.625

To find the angle theta, we need to take the inverse cosine (cos^-1) of 5.625:

theta = cos^-1(5.625)

Since the value of 5.625 is outside the range of -1 to 1, it is not a valid cosine value. Therefore, there is no real angle theta that satisfies the equation.

In this case, the magnitude of the vector product between vectors A and B would be zero, as there is no angle theta at which the sine function sin(theta) is defined.