Question

Vector A points to the north and has length A. Vector B points to the east and has length B = 2.0 A. Find the magnitude of C= 8.8A + B in terms of A.

Answer 1

magnitude = sqrt((8.88A)^2 + B^2)

= sqrt(78.854a^2 + A^2 )

= 8.9361A

Answer 2

First clean up your notation.

In vectors you want to find C = 3.6A + B; where Ay = A sin(90) and Bx = B cos(0) 90 deg is N and 0 deg is E. So we have Bx = 2A cos(0). Now find the XY components.

X: Cx = 3.6 Ax + Bx = 3.6 A cos(90) + 2A cos(0) = 2A
Y: Cy = 3.6 Ay + By = 3.6 A sin(90) + 2A sin(0) = 3.6A

So that:

C = sqrt(X^2 + Y^2) = sqrt(4A^2 + 3.6^2A^2) = A sqrt(4 + 3.6^2) = 4.118 A ANS. And, though you didn't ask, the angle of C re the X axis is theta = ATAN(3.6/2).

Answer 3

A=2B, so B=0.5A. C=8.8A+B, so C=8.8A+0.5A=9.3A.

Answer 4

A = Aj^

B = Bi^ = 2i^

C = 8.8A + B = 8.8Aj^ + 2i^

Magnitude of C = Sqrt ( 8.8*8.8*A^2 + 2*2)

Answer 5

C=8.8A +B

mag = sqrt(8.8^2+2^2) *A =9.02 A