Question

the scalar product of 2 vectors is 2√3 and the magnitude of their vector product is 2. find the angle between them​

Answer 1

Explanation:

According to the question,

∣

A

×

B

∣=

3

∣

A

.

B

∣

Expanding the vector and the dot products:

ABsinθ=

3

ABcosθ

⇒tanθ=

3

⇒θ=60

∘

=

3

π

Answer 2

Explanation:

Let θ be the angle between two vectors a⃗ & b⃗ then

The scalar product is given as

a⃗ ⋅b⃗ =23–√

|a⃗ ||b⃗ |cosθ=23–√ ......(1)

The magnitude of Vector product is given as

|a⃗ ×b⃗ |=2

|a⃗ ||b⃗ |sinθ=2 ......(2)

Dividing (2) by (1) as follows

|a⃗ ||b⃗ |sinθ|a⃗ ||b⃗ |cosθ=223√

tanθ=13√

tanθ=tan30∘

θ=30∘