Subtracting Vectors?

Question

Hey all, well i am in a bit trouble because i don't seem to understand vector subtraction. OK here it is:
V1 = -5i + 2j
V2 = -3i -9j

Now what i get is -8 and -7.But then i need to use the Pythagorean theorem to find sides on my triangle on the graph that i have. Since the number is negative i can't swuare root the last digit of the pothagorean theorem. WHat am i doing wrong,

Answer 1

To subtract one vector from another you just subtract the components

V1 - V2 = (-5-(-3))i + (2 - (-9))j = -2i + 11j

The numbers you gave (-8 and -7) would be the result of adding the vectors.

The components of the vector give the length of the 2 legs of a right triangle. The pythagorean theorem will give you the length of the hypotenuse which is the same as the magnitude of the vector.

Answer 2

I assume your question is to find the lengths of the three sides of the triangle. The vertices are at:

O(0, 0)
V1(-5, 2)
V2(-3, -9)

The length of:

OV1 = √[(-5)² + 2²] = √(25 + 4) = √29
OV2 = √[(-3)² + (-9)²] = √(9 + 81) = √90 = 3√10

V1V2 = √[(-5 + 3)² + (2 + 9)²] = √[(-2)² + 11²] = √(4 + 121)
V1V2 = √125 = 5√5

Answer 3

V1=-5i+2j,
V2=-3i-9j
V1-V2=-2i+11j
its magnitude=sqrt[-2)^2+11^2]
=sqrt[4+121
=sqrt(125)]
=11.18. if it makes an angle alpha with the negative X-axis,
tan(alpha)=11/2=5.5
alpha=79.70 degreesor 100.30 degrees with the positive X-axis.

Answer 4

V1 - V2 = (-5,+2) - (-3,-9) = (-2,+11)

Answer 5

You dont take the square root, you take the square.

8^2 + 7^2 = (ans)^2

Thats if your'e looking for the length of the hypotenus.