Question

Find the angle between two vectors $\vec a\ and \vec b$ with magnitudes $\sqrt{3}$ and 2 , respectively having $\vec a\ . \vec b = \sqrt{6}$

Answer 1

Let the two vectors be a and b.

Therefore,

|a| = √3 and |b| = 2

Now, there dot product is given,

|a.b| = √6

Using the formula,

|a.b| = |a||b|Cosθ

√6 = √3 × 2 Cosθ

∴ Cosθ = 1

Cosθ = Cos0°

Therefore,

θ = 0°

Thus, the angle between these vectors is zero degree.

Also, θ = 2πn ± α

Since, α = 0

Therefore, θ = 0, 2π, 4π, etc.

These are the possible angles between these two vectors.

Hope it helps.

Answer 2

$\bigstar$ Solution:

$[\vec{a}] = \sqrt{3}, \: [\vec{b}] = 2 \; and \; \vec{a}\: . \: \vec{b} = \sqrt{6}$

Now, we know that $\vec{a} \: . \: \vec{b} = [\vec{a}][\vec{b}] cos\: \theta$

∴ $\sqrt{6} = \sqrt{3} \times 2 \times cos \: \theta$

$cos \: \theta = \frac{\sqrt{6} }{\sqrt{3} \times 2 }$

$cos \: \theta = \frac{1}{\sqrt{2} }$

$\implies \theta=\frac{\pi }{4}$

Thus, the angle between the given vectors $\vec{a}$ and $\vec{b}$ is $\frac{\pi }{4}$