Question

Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vectors?

Answer 1

Answer:

Value of angle between vector a and b is $56.30^{\circ}$.

Explanation:

Vectors a and b have scalar product 6.00

Let $\theta$ be the angle between a and b.

$\vec{a}.\vec{b} = 6$

ab cos $\theta$ = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

$\vec{a} \times\vec{b} = 9$

ab sin $\theta$ = 9 ...(2)

Dividing equation (2) by (1) we get

$\frac{ab sin \theta}{ab cos \theta} = \frac{9}{6}$

tan $\theta$ = 1.5

$\theta = tan ^{-1} (1.5)$

$\theta$ = $56.30^{\circ}$

Thus, value of angle between vector a and b is $56.30^{\circ}$.