Question

The magnitude of vector product of two unit vectors making an angle of 60 with each other is

(A) 1 (B) 2 (C) 3/2. (D) 32

Answer 1

Question:

The magnitude of vector product of two unit vectors making an angle of 60 with each other is

• 1
• 2
• 3/2
• 32
• $\frac{ \sqrt{3}}{2}$

Solution:

We know that the magnitude of unit vector is always 1 as it only shows us the direction of the vector.

Let a vector be $\vec{A}$.

It's unit vector can be written as: $\frac{\vec{A}}{|A|} = \vec{ ‎ ‎}$

What is left with us is direction and magnitude 1.

So, as per the question,

Let two vectors be $\vec{A}$ and $\vec{B}$

Its unit vector will be $\hat{A}$ and $\hat{B}$

Vector product of $\hat{A}$ and $\hat{B}$ making an angle 60° :

$\hat{A} \times \hat{B} = \hat{A}\hat{B}sin60^{\circ} \: \: \: \: \bigg[sin60^{\circ} = \frac{ \sqrt{3}}{2} \bigg] \\\\ \hat{A} \times \hat{B} = 1 \times 1 \times \frac{ \sqrt{3}}{2} \\\\ \implies \frac{ \sqrt{3}}{2}$