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

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The discussion focuses on the conditions under which the ratio of two vectors, $\vec{A}$ and $\vec{B}$, in the xy-plane can be expressed as the ratio of their x-components. The analysis shows that the equality $\frac{|\vec{A}|}{|\vec{B}|} = \frac{A_x}{B_x}$ holds true only when the y-components of both vectors are zero. The mathematical derivation confirms that $\frac{A_y}{A_x} = \pm \frac{B_y}{B_x}$, leading to the conclusion that the vectors must be aligned along the x-axis for the ratios to be equal.

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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
 
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I'm going to assume we're talking about the ratio of magnitudes. Suppose:

$$\vec{A}=\left\langle A_x,A_y \right\rangle$$

$$\vec{B}=\left\langle B_x,B_y \right\rangle$$

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

$$\frac{A_x^2+A_y^2}{B_x^2+B_y^2}=\frac{A_x^2}{B_x^2}$$

$$A_x^2B_x^2+A_y^2B_x^2=A_x^2B_x^2+A_x^2B_y^2$$

$$A_y^2B_x^2=A_x^2B_y^2$$

$$\frac{A_y^2}{A_x^2}=\frac{B_y^2}{B_x^2}$$

$$\frac{A_y}{A_x}=\pm\frac{B_y}{B_x}$$

What conclusion may we draw from this result?
 
I would immediately have a problem with \frac{\vec{A}}{\vec{B}}. The division of vectors is not defined. Did you mean \frac{|\vec{A}|}{|\vec{B}|}? That would be equal to \frac{A_x}{B_x} if and only if the other components of \vec{A} and \vec{B} are 0.
 
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