Vectors graphing terminology convensions

[EulersNumber]

Hello,

I'm stuck on a word problem that states:

A vector B with a magnitude of 3.5 meters is added to vector A, which lies along the x-axis. The sum of these two vectors is a third vector that lies along the y-axis and has a magnitude twice the magnitude of vector A. What is the magnitude of vector A?

I've no idea what is meant by "lies along the x/y axis". It almost sounds like a unit vector or a vector (working with 2D Cartesian vectors in physics class for context) that is parallel to the x axis and another parallel to the y axis. This would be easier to solve as that would make vector B the hypotenuse of a right angle triangle. Except for other exercises where this hasn't apparently been the case. Am I overthinking it? What could they mean?

 

[mario99]

The main information we have is this:

[imath]\bold{A} + \bold{B} = \bold{C}[/imath]

[imath]A + B = C = 2A[/imath]

Magnitude of vector [imath]\bold{A} = |\bold{A}| = A = \sqrt{A_1^2 + A_2^2}[/imath]

Now break the vectors intor their components.

[imath]\bold{A} = A_1i + A_2j[/imath]

[imath]\bold{B} = B_1i + B_2j[/imath]

[imath]\bold{C} = C_1i + C_2j[/imath]

Add the [imath]x[/imath] components together and [imath]y[/imath] components together. Then replace whatever component with its given information.

 

[fresh_42]

A,B,C mean only the lengths of these vecors here. Of course, A and C point in two different directions, so the vector equation does not hold, i.e.[imath] \vec{C}\neq 2\vec{A}. [/imath] However, C is twice as long as A.

 

[mario99]

The main information we have is this:

[imath]\bold{A} + \bold{B} = \bold{C}[/imath]

[imath]A + B = C = 2A[/imath]

Correction
The third line should be written like this:
[imath]C = 2A[/imath]