4. The cross product can also be determined if we know the components of the two vectors involved. The easiest way to do this is to write the cross product the determinant of a 3x3 matrix. The first row of the matrix is the unit vectors i, j, and k. The 2nd row is the components of the first vector and the 3rd row is the components of the second vector. The determinant of the 3x3 matrix can be written in terms of determinants of 2x2 matrices: î Ĵ AxB = A. Ay Az|=1 |Bx By B₂ The determinant of a 2x2 matrix is the product of the diagonal terms minus the product of the off-diagonal terms. Therefore k A-¹ - + Ax Az Bx B₂ k b. What is the magnitude of the torque? Ax Ayl Bx Byl Ax B= i(A,B₂ - A₂By) - j(AxB₂ - A₂Bx) + k(AxBy – AyBx) a. Let F = (21-3k) m and F = (51-4j-10R) N. Determine the torque in component vector form. Don't forget to include units.

4. The cross product can also be determined if we know the components of the two vectors involved. The easiest way to do this is to write the cross product the determinant of a 3x3 matrix. The first row of the matrix is the unit vectors i, j, and k. The 2nd row is the components of the first vector and the 3rd row is the components of the second vector. The determinant of the 3x3 matrix can be written in terms of determinants of 2x2 matrices: î Ĵ AxB = A. Ay Az|=1 |Bx By B₂ The determinant of a 2x2 matrix is the product of the diagonal terms minus the product of the off-diagonal terms. Therefore k A-¹ - + Ax Az Bx B₂ k b. What is the magnitude of the torque? Ax Ayl Bx Byl Ax B= i(A,B₂ - A₂By) - j(AxB₂ - A₂Bx) + k(AxBy – AyBx) a. Let F = (21-3k) m and F = (51-4j-10R) N. Determine the torque in component vector form. Don't forget to include units.

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter14: Static Equilibrium, Elasticity, And Fracture

Section: Chapter Questions

Problem 8PQ

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