Find the angle between each of the following pairs of vectors A⃗ =Axi^+Ayj^ and B⃗ =Bxi^+Byj^?

Part A

Ax1= -1.00, Ay1= 4.80; Bx1= 3.00, By1= -3.40.

Part B

Ax2= 4.00, Ay2= 5.00; Bx2= 12.0, By2= 5.60.

Answer:

use this

tan(angle) = y / x

A

tan(angle)A = 4.80/-1.00 = -4.80

angle A = – 78.2 in 2nd quadrant or 101.8°

tan(angle) B = -3.40 / 3.00 = -48.6 ° in 4th quadrant

so

101.8 + 48.6 = -150.4° or about 150° = angle between A and B

B

(angle) A = arctan(5.00 / 4.00) = 51.3° in 1st quadrant

(angle) B = arctan(5.60 / 12.0) = 25.0° in 1st quadrant

so

51.3 – 25.0 = 26.3° between them

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