Question

The magnitude of scalar product of two vectors is 8 and that of vector product is 8 √3. The angle
between them is :
(B) 60°
(C) 120°
(D) 150°
(A) 30°
​

Answer 1

a•b = abcos x = 8

a×b = absin x = 8√3

absin x = √3 abcos x

sin x = √3 cos x

tan x = √3

x = tan ^-1 √3

x = 60°

Answer 2

The angle between two vectors = 60°

Given :

The magnitude of scalar product of two vectors is 8 and that of vector product is 8√3

To find :

The angle between two vectors is

(A) 30°

(B) 60°

(C) 120°

(D) 150°

Solution :

Step 1 of 2 :

Write down the given data

Let the given vectors are $\vec{a}$ and $\vec{b}$

Also let θ be the angle between two vectors

Since magnitude of scalar product of two vectors is 8

Thus we get

$| \vec{a}. \vec{b}| = 8$

$\implies | \vec{a} || \vec{b}| \cos \theta = 8 \: \: - - (1)$

Again magnitude of vector product of two vectors is 8√3

Thus we get

$| \vec{a} \times \vec{b}| = 8 \sqrt{3}$

$\implies | \vec{a} || \vec{b}| \sin \theta = 8 \sqrt{3} \: \: - - (2)$

Step 2 of 2 :

Find the angle between two vectors

From Equation 2 and Equation 1 we get

$\implies \: \dfrac{| \vec{a} || \vec{b}| \sin \theta}{| \vec{a} || \vec{b}| \cos \theta} = \dfrac{8 \sqrt{3} }{8}$

$\implies \tan \theta = \sqrt{3}$

$\implies \theta = {60}^{ \circ}$

So the angle between two vectors = 60°

Hence the correct option is (B) 60°

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