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Chapter 6: Problem 59

Two vectors have equal magnitude, and their scalar product is one third the square of their magnitude. Find the angle between them.

Short Answer

Expert verified
The angle between the two vectors is approximately 70.53 degrees or 1.23 radians

Step by step solution

01

Understand the problem

The formula for the dot product of two vectors is \( \vec{A} \cdot \vec{B} = ||\vec{A}|| \cdot ||\vec{B}|| \cdot cos(\Theta) \), where \( ||\vec{A}|| \) and \( ||\vec{B}|| \) are the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\), and \( \Theta \) is the angle between them. In this exercise, we're given that the magnitudes of vectors \(\vec{A}\) and \(\vec{B}\) are the same (let's denote it as M). We are also given information about their dot product which is equal to one third of the square of their magnitude \( \frac{M^{2}}{3} \).
02

Substitute given values into dot product formula

Substitute the provided information into the dot product formula: \( \frac{M^{2}}{3}= M \cdot M \cdot cos(\Theta)\). Simplifying this equation gives us \( cos(\Theta)= \frac{1}{3} \).
03

Inverse cosine

Finally, to solve for \( \Theta \), you can use the inverse cosine function (or arccos). The inverse cosine of \( \frac{1}{3} \) equals to \( \Theta \). Please remember that the arccos function returns a value in the range from 0 to \(\pi\) radians, which corresponds to the range from 0 to 180 degrees.
04

Calculate the angle

Calculating the arccos of \( \frac{1}{3} \) will provide the desired angle.

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