Question

if the scalar and vector product of two vector a and b are equal in magnitude then the angle between two vector is​

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

✼ Scalar product and vector product of two vectors A and B are equal in magnitude.

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

❅ We have to find angle between two vectors.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

Let two vectors A and B are inclined at angle Φ.

✠ Scalar (dot) product :

• A • B = AB cosΦ

✠ Vector (cross) product :

• A × B = AB sinΦ

ATQ,

➝ A • B = A × B

➝ AB cosΦ = AB sinΦ

➝ sinΦ/cosΦ = AB/AB

➝ tanΦ = 1

➝ Φ = tanˉ¹ (1)

➝ Φ = 45°

Answer 2

Given ,

The scalar and vector product of two vector a and b are equal in magnitude

As we know that , the scalar product and dot product of two vector is given by

$\boxed{\tt{a.b = ab \cos( \theta)}}$

And

$\boxed{\tt{a \times b = ab \sin( \theta)}}$

According to the question ,

$\tt \implies ab \cos( \theta) = ab \sin( \theta)$

$\tt \implies \frac{ \sin( \theta) }{ \cos( \theta) } = \frac{ab}{ab} \: \: or \: \: \frac{ \cos( \theta) }{ \sin( \theta) } = \frac{ab}{ab}$

$\tt \implies \tan( \theta) = 1 \: \: or \: \: \cot( \theta) = 1$

$\tt \implies \theta = 45$

• The angle between given two vector is 45