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

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The discussion centers on the conditions under which the ratio of two vectors, A/B, equals the ratio of their x-components, A_x/B_x. It is established that if A and B are defined in the xy-plane, the equality holds when the y-components of both vectors are zero. The mathematical derivation shows that the ratios of the y-components to the x-components must be equal, leading to the conclusion that A_y/A_x equals B_y/B_x. The conversation highlights the confusion around vector division, clarifying that the intended interpretation involves the magnitudes of the vectors rather than direct vector division. Thus, the condition for the ratios to be equal is that the y-components must be zero.
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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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