Which vector has a magnitude to the nearest tenth of 3.6?

Question

A.<4,-2>

B.<1,3>

C.<3,2>

D.<4,2>

Answer 1

The vectors are coordinates with the origin <0,0> as the starting point of the vector.  We can use the Pythagorean theorem to find the magnitude of a vector.  Lets test out each answer choice.

 

c = √(x² + y²)

 

where c is magnitude of vector.

 

For choice A

 

c = √(4² + (-2)²)

c = √(16 + 4)

c = 4.8

 

 

For choice B

 

c = √(1² + 3²)

c = √(1 + 9)

c = 3.2

 

 

For choice C

 

c = √(3² + 2²)

c = √(9 + 4)

c = 3.6

 

 

Choice D is the same as choice A.  Only that the coordinates of x and y are flipped.

 

The only answer choice in which the vector is or comes close to 3.6 is choice C.