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Chapter 3: Problem 14

Vector \(\overrightarrow{\mathbf{A}}\) has magnitude 4.0 units; vector \(\overrightarrow{\mathbf{B}}\) has magnitude 6.0 units. The angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\) is \(60.0^{\circ} .\) What is the magnitude of \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}} ?\)

Short Answer

Expert verified
Answer: The magnitude of the sum \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) is approximately \(8.72\) units.

Step by step solution

01

Understand the Law of Cosines for Vectors

The Law of Cosines for vectors states that the magnitude of the sum of two vectors can be calculated using the magnitudes of the vectors and the angle between them. The formula is: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{||\overrightarrow{\mathbf{A}}||^{2} + ||\overrightarrow{\mathbf{B}}||^{2} + 2||\overrightarrow{\mathbf{A}}||||\overrightarrow{\mathbf{B}}||\cos{\theta}}$$ Where \(||\overrightarrow{\mathbf{A}}||\) is the magnitude of vector \(\overrightarrow{\mathbf{A}}\), \(||\overrightarrow{\mathbf{B}}||\) is the magnitude of vector \(\overrightarrow{\mathbf{B}}\), and \(\theta\) is the angle between \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\).
02

Insert the given values into the formula

Now you can insert the given values into the Law of Cosines formula: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{4^{2} + 6^{2} + 2(4)(6)\cos{60^{\circ}}}$$
03

Calculate the cosine of the angle

First, calculate the cosine of the given angle: $$\cos{60^{\circ}} = 0.5$$
04

Perform the calculation

Now that you have the cosine value, substitute it back into the formula and perform the calculation: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{4^{2} + 6^{2} + 2(4)(6)(0.5)}$$ Simplify the expression: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{16 + 36 + 24}$$ $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{76}$$
05

Final Answer

Therefore, the magnitude of the sum \(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}\) is: $$||\overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{B}}|| = \sqrt{76} \approx 8.72$$ units.

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