Question

The magnitude of two vectors are4and 6and the magnitude of their scalar product is 12√2 then the angle between the these vectors is?

Answer 1

Answer:

45°

Explanation:

let,

|A| = 4

|B| = 6

Given that:

|A|.|B| = 12$\sqrt{2}$ ----->(1)

scalar product Or dot product,

|A|.|B| = |A| |B| cosθ ----->(2)

From (1) and (2) :

=> |A| |B| cosθ = 12$\sqrt{2}$

=> 4 × 6 cosθ = 12$\sqrt{2}$

=> 24 cosθ = 12$\sqrt{2}$

=> cosθ = $\frac{12\sqrt{2}}{24}$

=> cosθ = $\frac{1}{\sqrt{2}}$

=> cosθ = cos 45°

=> $\fbox{ θ = 45° }$