Question

Exercise 1.48 13 of 13 PartA For the two vectors A and B in the igure (Figure 1), find the scalar product A-B A- B Submit Request Answer Part B Find the magnitude of the vector product A x B Submit Request Answer Part C What is the direction of the vector product A x B Figure 1 of 1 A (3.60 m) Correct < Return to Assignment Provide Feedback 70.0° 300 0 B (2.40 m)

Answer 1

A vector is a physical quantity having magnitude as well as direction and which can be added according to the triangle rule. To find the scalar product and vector product first we will find the components of given vector and then their product.

Calculation work is shown below.

A vector is a physical quantity having magnitude as well as direction and which can be added according to the triangle rule. To find the scalar product and vector product first we will find the components of given vector and then their product. Calculation work is shown below. Given vector is as shown in figure A^vector = 3.60 cos 70 i^vector + 3.60 sin 70 j^vector = 1.23i^vector + 3.38 J^vector B^vector = 2.40 cos 30(-i^vector) + 2.40 sin 30 (-j^vector) = -2.07 i^vector - 1.2 J^vector B^vector = -(2.07i^vector + 1.2J^vector) Part A: Scalar product A^vector middot B^vector: (1.23i^vector + 3.38 j^vector) use: i^vector middot i^vector = J = j^vector middot J^vector = k^vector middot k^vector = 1 i^vector middot J^vector = j^vector middot k^vector = k^vector middot i^vector = 0 A^vector middot B^vector = 1.23 times (-2.07) + 3.38 times (-1.2) = -2.546 + (-4.056) = -6.602 A^vector middot B^vector = -6.602 part B vector product