/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Two vectors \(\mathbf{A}\) and \... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 51

Two vectors \(\mathbf{A}\) and \(\mathbf{B}\) have precisely equal magnitudes. In order for the magnitude of \(\mathbf{A}+\mathbf{B}\) to be one hundred times larger than the magnitude of \(\mathbf{A}-\mathbf{B},\) what must be the angle between them?

Short Answer

Expert verified
The angle between vectors \(\mathbf{A}\) and \(\mathbf{B}\) must be approximately 1.05 degrees.

Step by step solution

01

Inserting the magnitude relations into equations

From \(|\mathbf{A} + \mathbf{B}| = 100\times (\mathbf{A} - \mathbf{B}|)\), we can get \(\sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 + 2|\mathbf{A}||\mathbf{B}|\cos(\theta)} = 100 \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2 - 2|\mathbf{A}||\mathbf{B}|\cos(\theta)}\), since \(|\mathbf{A}| = |\mathbf{B}|\), we can simplify the equation into \(\sqrt{4|\mathbf{A}|^2 + 2|\mathbf{A}||\mathbf{B}| \cos(\theta)} = 100 \sqrt{2|\mathbf{A}|^2 - 2|\mathbf{A}|^2\cos(\theta)}\).
02

Simplifying the equation

Squaring both sides to remove the square roots: \(4|\mathbf{A}|^2 + 2|\mathbf{A}|^2 \cos(\theta) = 10000 |\mathbf{A}|^2 - 10000 |\mathbf{A}|^2 \cos(\theta)\), Divide both sides by \(2|\mathbf{A}|^2\): \(2 + \cos(\theta) = 5000 - 5000 \cos(\theta)\).
03

Solving for the angle

Moving terms to get all \(\cos(\theta)\) on one side, yields: \(5001 \cos(\theta) = 4998\). Dividing both sides by 5001 gives \(\cos(\theta) = 4998 / 5001 \approx 0.9994\). Taking the arccosine of both sides will give us the value of the angle \(\theta\).
04

Calculating the angle

Using a scientific calculator, we find that \(\arccos(0.9994)\) is approximately 1.05 degrees.

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