Two vectors possess magnitudes of 2 m and 3 m . The angle between them is 60^ ^ ^ . Determine the magnitude of their vector product.

Correct answer: C. 3 3 m^2 The magnitude of the vector product of two vectors A and B is given by | A × B | = | A | | B | θ . Given | A | = 2 m , | B | = 3 m , and θ = 60^ ^ ^ . Substituting the values: | A × B | = 2 × 3 × 60^ ^ ^ | A × B | = 6 × 3 2 = 3 3 m^2

Concepts Of Physics MCQ Edition [Volume 1]

Exam: Concepts Of Physics MCQ Edition [Volume 1]
Subject: Physics
Chapter: Physics and Mathematics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Physics and Mathematics

Two vectors possess magnitudes of 2 m and 3 m . The angle between...

Two vectors possess magnitudes of

2 \text{ m}

and

3 \text{ m}

. The angle between them is

60^{}^{{}^{\circ}}

. Determine the magnitude of their vector product.

Options

A $6\sqrt{3} \text{ m}^2$
B $3 \text{ m}^2$
C $3\sqrt{3} \text{ m}^2$
D $6 \text{ m}^2$

Correct answer

C. $3\sqrt{3} \text{ m}^2$

Step-by-step solution

The magnitude of the vector product of two vectors

\vec{A}

and

\vec{B}

is given by

|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta

.

Given

|\vec{A}| = 2 \text{ m}

|\vec{B}| = 3 \text{ m}

, and

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

.

Substituting the values:

|\vec{A} \times \vec{B}| = 2 \times 3 \times \sin 60^{}^{{}^{\circ}}

|\vec{A} \times \vec{B}| = 6 \times \dfrac{\sqrt{3}}{2} = 3\sqrt{3} \text{ m}^2

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