Question

if the magnitude of two vectors are 3 and 4 and magnitude of scalar product is 6.what is the agle between the vectors?​

Answer 1

Let the two vectors be $\sf{\vec{x} \ and \ \vec{y}}$

Given,

• Their magnitudes are 3 and 4 respectively

• Their scalar product (dot product) is 6

To finD

Angle between the two vectors

Dot Product

$\displaystyle{\sf{ \vec{x} . \vec{y} = | \vec{x} | \times | \vec{y}| \times \cos( \alpha ) }} \\ \\ \longrightarrow \: \sf{6 = 3 \times 4 \times \cos( \alpha ) } \\ \\ \longrightarrow \: \sf{ \cos( \alpha ) = \frac{1}{2} } \\ \\ \large{ \longrightarrow \: \boxed{ \boxed{ \sf{ \alpha = {60}^{ \circ} }}}}$

The angle between the vectors would be 60°

Answer 2

Magnitude of vectors = 3 and 4

Scalar Product  = 6

Find the angle between the vectors using general formula.

$\cos\theta = \dfrac{a.b}{|a||b|}\\\\cos \theta = \dfrac{6}{3\times4}\\\\\cos \theta = \dfrac{1}{2}\\\\\boxed{\theta=60^o}$

Hence, angle between the vectors = 60°.