Question

Vector → A has a magnitude of 3.7 units, and vector → B has a magnitude of 11.5 units. If the value of → A ⋅ → B is 22.4 units squared, what is the angle (in degrees) between the two vectors?

Answer 1

Answer:

$\theta=58.26^{\circ}$

Explanation:

It is given that,

Magnitude of vector A, $|A|=3.7\ units$

Magnitude of vector B, $|B|=11.5\ units$

Dot product of two vectors, $A.B=22.4\ unit^2$

Let $\theta$ is the angle between the two vectors. We know that the angle between two vectors is given by the formula of dot product as :

$A.B=|A||B|\ cos\theta$

$cos\tehta=\dfrac{A.B}{|A||B|}$

$cos\tehta=\dfrac{22.4}{3.7\times 11.5}$

$cos\theta=0.526$

$\theta=58.26^{\circ}$

So, the angle between two vectors is 58.26 degrees. Hence, this is the required solution.