Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Point \(A(-4,2,5)\) and the two vectors, \(\mathbf{R}_{A M}=(20,18-10)\) and \(\mathbf{R}_{A N}=(-10,8,15)\), define a triangle. Find \((a)\) a unit vector perpendicular to the triangle; \((b)\) a unit vector in the plane of the triangle and perpendicular to \(\mathbf{R}_{A N} ;(c)\) a unit vector in the plane of the triangle that bisects the interior angle at \(A\).

Write an expression in rectangular components for the vector that extends from \(\left(x_{1}, y_{1}, z_{1}\right)\) to \(\left(x_{2}, y_{2}, z_{2}\right)\) and determine the magnitude of this vector.

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 \(\mathbf{E}=0 .\)

Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft. We assume a constant altitude, a plane earth, a flight along the \(x\) axis from 0 to 10 units, no vertical velocity component, and no change in wind velocity with time. Assume \(\mathbf{a}_{x}\) to be directed to the east and \(\mathbf{a}_{y}\) to the north. The wind velocity at the operating altitude is assumed to be: $$ \mathbf{v}(x, y)=\frac{\left(0.01 x^{2}-0.08 x+0.66\right) \mathbf{a}_{x}-(0.05 x-0.4) \mathbf{a}_{y}}{1+0.5 y^{2}} $$ Determine the location and magnitude of \((a)\) the maximum tailwind encountered; \((b)\) repeat for headwind; \((c)\) repeat for crosswind; \((d)\) Would more favorable tailwinds be available at some other latitude? If so, where?

Two unit vectors, \(\mathbf{a}_{1}\) and \(\mathbf{a}_{2}\), lie in the \(x y\) plane and pass through the origin. They make angles \(\phi_{1}\) and \(\phi_{2}\), respectively, with the \(x\) axis \((a)\) Express each vector in rectangular components; \((b)\) take the dot product and verify the trigonometric identity, \(\cos \left(\phi_{1}-\phi_{2}\right)=\cos \phi_{1} \cos \phi_{2}+\sin \phi_{1} \sin \phi_{2} ;(c)\) take the cross product and verify the trigonometric identity \(\sin \left(\phi_{2}-\phi_{1}\right)=\sin \phi_{2} \cos \phi_{1}-\cos \phi_{2} \sin \phi_{1}\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free