Question 94: Motion in 2D & Vectors

Question

Vector $\vec{A}$ of magnitude $5 \sqrt{3}$ units, another vector $\vec{B}$ of magnitude of 10 units are inclined to each other at an angle of $30^{\circ}$. The magnitude of vector product of the two vectors is $\left[\sin 30^{\circ}=\frac{1}{2}\right]$

Accepted Answer

We have
$ |\vec{A}| = 5\sqrt{3}, \quad |\vec{B}| = 10, \quad 	heta = 30^\circ. $
The magnitude of the cross product is
$ |\vec{A} 	imes \vec{B}| = |\vec{A}| \,|\vec{B}| \sin	heta. $
Substitute the values:
$ |\vec{A} 	imes \vec{B}| = (5\sqrt{3})(10)(\sin 30^\circ). $
$ = 50\sqrt{3} \cdot \frac{1}{2} $
$ = 25\sqrt{3}. $
Ã¢ÂÂ Correct Answer:
Option C: $25 \sqrt{3}$ units

Answer

$5 \sqrt{3}$ units

Answer

10 units

Answer

$25 \sqrt{3}$ units

Answer

75 units

Motion in 2D & Vectors

Vector \vec{A} of magnitude 5 \sqrt{3} units, another vector \vec{B} of magnitude of 10 units are inclined to each other at an angle of 30^{\circ}. The magnitude of vector product of the two vectors is \left[\sin 30^{\circ}=\frac{1}{2}\right]

Choose the correct answer: