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Physics 71 Under what conditions does the ratio A}/B equal A_x//B_x

karush

Well-known member
Jan 31, 2012
2,724
71.15 Two vectors $\vec{A}$ and $\vec{B}$ lie in xy plane.
Under what conditions does the ratio $\vec{A}/\vec{B}$ equal $\vec{A_x}/\vec{B_x}$?

Sorry but I had a hard time envisioning what this would be???
also thot I posted this earlier but I can't find it
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I'm going to assume we're talking about the ratio of magnitudes. Suppose:

\(\displaystyle \vec{A}=\left\langle A_x,A_y \right\rangle\)

\(\displaystyle \vec{B}=\left\langle B_x,B_y \right\rangle\)

Then, let's see what happens when we write:

\(\displaystyle \frac{A_x^2+A_y^2}{B_x^2+B_y^2}=\frac{A_x^2}{B_x^2}\)

\(\displaystyle A_x^2B_x^2+A_y^2B_x^2=A_x^2B_x^2+A_x^2B_y^2\)

\(\displaystyle A_y^2B_x^2=A_x^2B_y^2\)

\(\displaystyle \frac{A_y^2}{A_x^2}=\frac{B_y^2}{B_x^2}\)

\(\displaystyle \frac{A_y}{A_x}=\pm\frac{B_y}{B_x}\)

What conclusion may we draw from this result?
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I would immediately have a problem with [tex]\frac{\vec{A}}{\vec{B}}[/tex]. The division of vectors is not defined. Did you mean [tex]\frac{|\vec{A}|}{|\vec{B}|}[/tex]? That would be equal to [tex]\frac{A_x}{B_x}[/tex] if and only if the other components of [tex]\vec{A}[/tex] and [tex]\vec{B}[/tex] are 0.
 
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