Given two vectors a = 3.6 i -3.9 j and b =6.5 i +8.9 j what is the magnitude (in

ID: 1593491 • Letter: G

Question

Given two vectors a = 3.6i -3.9j and b =6.5i+8.9j
what is the magnitude (in m) of the vector a?

From the information given above, find the following and show all work:

1) Find the direction (in ° = deg) of the vector a

2) Find the magnitude of the vector b.

3) Find the direction (in °) of the vector b.

4) Find the magnitude of the vector a+b.

5) Find the direction of the vector a+b.

6) Find the magnitude of the vector b-a.

7) Find the direction of the vector b-a

8) Find the magnitude of the vector a-b.

9) Find the direction of the vector a-b

magnitude of a = sqrt (ax2 +ay2) = sqrt (28.17) = 5.31m

1) direction of vector a = tan-1 (ay/ ax)= tan-1 (-3.9/3.6) = - 47.29°

2) magnitude of b = sqrt (bx2 +by2) = sqrt (121.46) = 11.02m

3) direction of vector b = tan-1 (by/ bx)= tan-1 (8.9/6.5) = 53.86°

4) Vector a + b = 10.1i+5j

Magnitude of a +b = sqrt ((a+b)x2 + (a+b)y2 ) = sqrt (127.01) = 11.26 m

5) Direction of vector a+b = tan-1 ((a+b)y/ (a+b)x)= tan-1 (5/10.1) = 26.34°

6) Vector b - a = 2.9i+12.8j

Magnitude of b - a = sqrt ((b-a)x2 + (b-a)y2 ) = sqrt (172.25) = 13.12 m

7) Direction of vector b -a = tan-1 ((b-a)y/ (b -a)x)= tan-1 (12.8/2.9) = 77.23°

8) Vector a-b = -2.9 i-12.8j

Magnitude of a - b = sqrt ((a- b)x2 + (a- b)y2 ) = sqrt (172.25) = 13.12m

9) Direction of vector a-b = 180 + tan-1 ((a-b)y/ (a-b)x)= 180 + tan-1 (12.8/2.9) = 180 + 77.23 = 257.23°

(if both component of vector is -ive than the vector lies in 3rd quadrant, so the angle will be 180 +... )

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