Vector Algerbra and Geometry help

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AlphaThrone

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I don't understand what exactly K is in the red circle. How is it a constant if it's equal to 2n3/W and n1/H? Any help is appreciated.
 
AlphaThrone said:
View attachment 33389
I don't understand what exactly K is in the red circle. How is it a constant if it's equal to 2n3/W and n1/H? Any help is appreciated.
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It looks to me like H and W are also constants. The n's are merely the magnitudes of unit vectors, if I am reading the problem correctly.

It would help if you would show us Q1.d as well, for reference.

-Dan
 
topsquark said:
It looks to me like H and W are also constants. The n's are merely the magnitudes of unit vectors, if I am reading the problem correctly.

It would help if you would show us Q1.d as well, for reference.

-Dan
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1657734775690.png
Sorry, here is the full question. I get what H and W are, I just don't understand where the K comes in and what exactly it represents.
 
I am assuming your question is about [imath]k[/imath], i.e., low-case, not [imath]K[/imath]. If you have an equation with two variables of the type [imath]An_1 = Bn_3[/imath] then there are infinitely many solutions, but they all look like [imath]n_1 = kB, n_3 = kA[/imath], where [imath]k[/imath] is initially an arbitrary number. But since you are looking for a unit vector you need [imath]n_1^2+n_3^2 = 1[/imath], i.e. [imath]k=\frac{1}{A^2+B^2}[/imath].
 
blamocur said:
I am assuming your question is about [imath]k[/imath], i.e., low-case, not [imath]K[/imath]. If you have an equation with two variables of the type [imath]An_1 = Bn_3[/imath] then there are infinitely many solutions, but they all look like [imath]n_1 = kB, n_3 = kA[/imath], where [imath]k[/imath] is initially an arbitrary number. But since you are looking for a unit vector you need [imath]n_1^2+n_3^2 = 1[/imath], i.e. [imath]k=\frac{1}{A^2+B^2}[/imath].
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Ooops -- done it again :(: it's [imath]k=\frac{1}{\sqrt{A^2+B^2}}[/imath]
 
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