For the vectors in Fig. E1.28, (a) Find the magnitude and direction of the vector product (b) find the magnitude and direction of
Step 1 of 3
Solution 49E Part 1: Here the magnitudes of the vectors are, | | 8 m | | 10 m. The angle between these two vectors is, 0 0 0 0 90 + (90 53 ) = 127 . The cross product of these vectors will be, 0 A × D = A| || | = A D| || |27 = 8 × 10 × 0.972 = 77.81 CONCLUSION: So, the magnitude of A × D is 77.81 meters . The direction is going into the page. By the right hand screw rule, the cross product always gives a vector whose direction is perpendicular to the plane where the two vectors lie. Part 2: D × A = (A × D) = 77.81 meters The magnitude is the same as A × D, but the direction is opposite. The minus sign here tells about the direction of D × A. It will be coming out of the page.
Textbook: University Physics
Author: Hugh D. Young, Roger A. Freedman
The full step-by-step solution to problem: 49E from chapter: 1 was answered by , our top Physics solution expert on 05/06/17, 06:07PM. This full solution covers the following key subjects: direction, Find, magnitude, fig, Product. This expansive textbook survival guide covers 26 chapters, and 2929 solutions. Since the solution to 49E from 1 chapter was answered, more than 707 students have viewed the full step-by-step answer. University Physics was written by and is associated to the ISBN: 9780321675460. This textbook survival guide was created for the textbook: University Physics, edition: 13. The answer to “For the vectors in Fig. E1.28, (a) Find the magnitude and direction of the vector product (b) find the magnitude and direction of” is broken down into a number of easy to follow steps, and 23 words.