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Vector A⃗ has magnitude 13.0 m and vector B⃗ has magnitude 16.0 m . The scalar product A⃗ ⋅B⃗ is 71.0 m2 .

What is the magnitude of the vector product between these two vectors?

1 Answer

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Final answer:

The magnitude of the vector product between vectors A and B is 197.12 m.

Step-by-step explanation:

The magnitude of the vector product between vectors A and B can be calculated using the formula:

A × B = |A| |B| sin(θ)

Given that the scalar product, A · B, is 71.0 m2 and the magnitudes of vectors A and B are 13.0 m and 16.0 m respectively, we can solve for sin(θ) using the equation:

A · B = |A| |B| cos(θ)

Substituting the given values into the equation, we have:

71.0 m2 = (13.0 m) (16.0 m) cos(θ)

cos(θ) = 71.0 m2 / (13.0 m × 16.0 m) = 71.0 m2 / 208.0 m2 = 0.341

Since cos(θ) = adjacent / hypotenuse, we can use the Pythagorean identity sin2(θ) + cos2(θ) = 1 to solve for sin(θ):

sin(θ) = sqrt(1 - cos2(θ)) = sqrt(1 - 0.3412) = sqrt(0.884) = 0.940

Finally, we can calculate the magnitude of the vector product:

|A × B| = |A| |B| sin(θ) = (13.0 m) (16.0 m) (0.940) = 197.12 m

User Sumit Chourasia
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