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Chapter 1: Problem 46

If \(|\vec{A}+\vec{B}|=|\vec{A}|=|\vec{B}|\), then the angle between \(\vec{A}\) and \(\vec{B}\) is (A) \(120^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(0^{\circ}\)

Short Answer

Expert verified
The angle between the vectors \(\vec{A}\) and \(\vec{B}\) is (A) \(120^{\circ}\).

Step by step solution

01

Cosine formula for finding the angle between two vectors

To find the angle between two vectors \(\vec{A}\) and \(\vec{B}\), we can use the cosine formula that involves the dot product and the magnitudes of the vectors: \[cos(\theta) = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|}\] where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\), and \(\vec{A}\cdot\vec{B}\) denotes their dot product.
02

Write down the equation for the sum of the vectors

We're given that the magnitudes of \(\vec{A}\), \(\vec{B}\) and \(\vec{A}+\vec{B}\) are equal, therefore: \[|\vec{A}+\vec{B}|^2=|\vec{A}|^2=|\vec{B}|^2\]
03

Expand the equation for the magnitude of the sum

We know that for any vector \(\vec{V}\), its squared magnitude can be denoted as: \[|\vec{V}|^2 =\vec{V} \cdot \vec{V} \] Now expand the magnitude of the sum: \[|\vec{A}+\vec{B}|^2=(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})\]
04

Distribute the dot products

Now distribute the dot product of the sum of the vectors: \[(\vec{A}+\vec{B})\cdot(\vec{A}+\vec{B})=\vec{A}\cdot\vec{A}+\vec{A}\cdot\vec{B}+\vec{B}\cdot\vec{A}+\vec{B}\cdot\vec{B}\] Remember that the dot product is commutative, so \(\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}\). Thus, we may rewrite the above equation as: \[\vec{A}\cdot\vec{A}+2\vec{A}\cdot\vec{B}+\vec{B}\cdot\vec{B}=|\vec{A}|^2+2\vec{A}\cdot\vec{B}+|\vec{B}|^2\]
05

Use the given condition to find angle

We know that \(|\vec{A}+\vec{B}|^2=|\vec{A}|^2=|\vec{B}|^2\). Substitute this in the equation we found in step 4: \[|\vec{A}|^2+2\vec{A}\cdot\vec{B}+|\vec{B}|^2=|\vec{A}|^2\] Now, since \(|\vec{A}|=|\vec{B}|\), we can write \(|\vec{A}|^2=|\vec{B}|^2\). Our equation becomes: \[|\vec{A}|^2+2\vec{A}\cdot\vec{B}+|\vec{A}|^2=|\vec{A}|^2\] Cancel out the terms, and we get: \[\vec{A}\cdot\vec{B}=-\frac{1}{2}|\vec{A}|^2\] Now use the cosine formula to get: \[cos(\theta) = \frac{-\frac{1}{2}|\vec{A}|^2}{|\vec{A}| \cdot |\vec{A}|}\] This simplifies to: \[cos(\theta) = -\frac{1}{2}\]
06

Find the angle

From the cosine function, we can find the angle: \[\theta=cos^{-1}\left(-\frac{1}{2}\right)\] and \[\theta = 120^{\circ}\] Therefore, the angle between \(\vec{A}\) and \(\vec{B}\) is: (A) \(120^{\circ}\)

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Most popular questions from this chapter

If \(|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}\), then the value of \(|\vec{A}+\vec{B}|\) is (A) \(\left(A^{2}+B^{2}+A B\right)^{1 / 2}\) (B) \(\left(A^{2}+B^{2}+\frac{A B}{\sqrt{3}}\right)^{1 / 2}\) (C) \((A+B)\) (D) \(\left(A^{2}+B^{2}+\sqrt{3} A B\right)^{1 / 2}\)

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