Question

A vector \(\vec{A}\) of magnitude \(7\sqrt{2}\) unit, another vector \(\vec{B}\) of magnitude \(14\) unit are inclined at an angle of \(60^{\circ }\). The magnitude of the scalar product \((\vec{A}\cdot \vec{B})\) is \(\left[\cos (60)^{\circ }=\frac{1}{2}\right]\)

• A. \(49\sqrt{2}\)
• B. \(7\sqrt{2}\)
• C. \(14\sqrt{2}\)
• D. \(9\sqrt{2}\)

Hint:

The scalar product equals the product of the two magnitudes and the cosine of the angle between them.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The scalar (dot) product of two vectors is \(\vec{A}\cdot\vec{B}=AB\cos\theta\), where \(\theta\) is the angle between them.

Step 2: Given Data
\(A=7\sqrt{2}\), \(B=14\), \(\theta=60^{\circ}\) and \(\cos 60^{\circ}=\frac{1}{2}\).

Step 3: Calculation
\(\vec{A}\cdot\vec{B}=7\sqrt{2}\times 14\times\frac{1}{2}=7\sqrt{2}\times 7=49\sqrt{2}\).

Step 4: Compare the options
This matches option (A).

Final Answer:
The magnitude of the scalar product is \(49\sqrt{2}\). \[ \boxed{\text{(A) }49\sqrt{2}} \]