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Chapter 1: Problem 8

Demonstrate the ambiguity that results when the cross product is used to find the angle between two vectors by finding the angle between \(\mathbf{A}=3 \mathbf{a}_{x}-2 \mathbf{a}_{y}+4 \mathbf{a}_{z}\) and \(\mathbf{B}=2 \mathbf{a}_{x}+\mathbf{a}_{y}-2 \mathbf{a}_{z} .\) Does this ambiguity exist when the dot product is used?

Short Answer

Expert verified
Answer: By using the dot product method, we can resolve the ambiguity in finding the angle between two vectors A and B, as the cosine function gives a unique angle within the range of [0, π].

Step by step solution

01

Calculate the cross product of A and B

The cross product of A and B can be found as: \(\mathbf{A} \times \mathbf{B} = ((-2)(-2) - (4)(1))\mathbf{a}_{x} + ((4)(2) - (3)(-2))\mathbf{a}_{y} + ((3)(1) - (2)(-2))\mathbf{a}_{z}\) Calculating the values, we get: \(\mathbf{A} \times \mathbf{B} = 4\mathbf{a}_{x} + 14\mathbf{a}_{y} + 7\mathbf{a}_{z}\)
02

Calculate the magnitudes of A, B, and AxB

To find the angle, we need the magnitudes of A, B, and AxB: \(|\mathbf{A}| = \sqrt{(3^2) + (-2^2) + (4^2)} = \sqrt{29}\) \(|\mathbf{B}| = \sqrt{(2^2) + (1^2) + (-2^2)} = \sqrt{9} = 3\) \(|\mathbf{A} \times \mathbf{B}| = \sqrt{(4^2) + (14^2) + (7^2)} = \sqrt{249}\)
03

Find the angle between A and B using cross product

Using the formula for the angle between two vectors using the cross product: \(\sin{\theta} = \frac{|\mathbf{A} \times \mathbf{B}|}{|\mathbf{A}| \cdot |\mathbf{B}|}\) \(\sin{\theta} = \frac{\sqrt{249}}{\sqrt{29} \cdot 3}\) Now, we can see the ambiguity in the cross-product method as the sine function has multiple solutions for an angle, usually leading to two possible angles. Therefore, we need to resolve this ambiguity by finding a unique angle.
04

Calculate the dot product of A and B

The dot product of A and B is given by, \(\mathbf{A} \cdot \mathbf{B} = (3)(2) + (-2)(1) + (4)(-2) = 6 - 2 - 8 = -4\)
05

Find the angle between A and B using dot product

Using the formula for the angle between two vectors using the dot product: \(\cos{\theta} = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| \cdot |\mathbf{B}|}\) \(\cos{\theta} = \frac{-4}{\sqrt{29} \cdot 3}\) Now, we can clearly see there's no ambiguity with the dot product. The cosine function gives a unique angle as the output, with a range of \([0, \pi]\). So, we can confirm that the ambiguity does not exist when using the dot product to find the angle between two vectors.

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Most popular questions from this chapter

Given the vector field \(\mathbf{E}=4 z y^{2} \cos 2 x \mathbf{a}_{x}+2 z y \sin 2 x \mathbf{a}_{y}+y^{2} \sin 2 x \mathbf{a}_{z}\) for the region \(|x|,|y|\), and \(|z|\) less than 2, find \((a)\) the surfaces on which \(E_{y}=0 ;(b)\) the region in which \(E_{y}=E_{z} ;(c)\) the region in which \(\mathbf{E}=0 .\)

Find the acute angle between the two vectors \(\mathbf{A}=2 \mathbf{a}_{x}+\mathbf{a}_{y}+3 \mathbf{a}_{z}\) and \(\mathbf{B}=\mathbf{a}_{x}-3 \mathbf{a}_{y}+2 \mathbf{a}_{z}\) by using the definition of \((a)\) the dot product; \((b)\) the cross product.

A sphere of radius \(a\), centered at the origin, rotates about the \(z\) axis at angular velocity \(\Omega \mathrm{rad} / \mathrm{s}\). The rotation direction is clockwise when one is looking in the positive \(z\) direction. ( \(a\) ) Using spherical components, write an expression for the velocity field, \(\mathbf{v}\), that gives the tangential velocity at any point within the sphere; \((b)\) convert to rectangular components.

Vector A extends from the origin to \((1,2,3)\), and vector \(\mathbf{B}\) extends from the origin to \((2,3,-2)\). Find \((a)\) the unit vector in the direction of \((\mathbf{A}-\mathbf{B})\); (b) the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of \(\mathbf{A}\) and \(\mathbf{B}\).

The vector from the origin to point \(A\) is given as \((6,-2,-4)\), and the unit vector directed from the origin toward point \(B\) is \((2,-2,1) / 3\). If points \(A\) and \(B\) are ten units apart, find the coordinates of point \(B\).
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