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

Correct answer: B. 3 m^2 Given | A | = 2 m and | B | = 3 m The angle between the vectors is θ = 60^ ^ ^ The scalar product of two vectors is given by A · B = | A | | B | θ Substituting the given values: A · B = 2 × 3 × 60^ ^ ^ A · B = 6 × (1)/(2) = 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 scalar product of the two vectors.

Options

A $6 \text{ m}^2$
B $3 \text{ m}^2$
C $1.5 \text{ m}^2$
D $3\sqrt{3} \text{ m}^2$

Correct answer

B. $3 \text{ m}^2$

Step-by-step solution

Given

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

and

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

The angle between the vectors is

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

The scalar product of two vectors is given by

\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta

Substituting the given values:

\vec{A} \cdot \vec{B} = 2 \times 3 \times \cos 60^{}^{{}^{\circ}}

\vec{A} \cdot \vec{B} = 6 \times \dfrac{1}{2} = 3 \text{ m}^2

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