Understanding Perpendicularity in a Plane

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SUMMARY

The discussion centers on the concept of perpendicularity in a plane, specifically regarding the line equation ax + by = 0 and its relationship with vectors A = [a, b] and B = [x, y]. It is established that vector A is perpendicular to line ax + by = 0 because the slopes of A and the line are negative reciprocals, confirming that A.B = 0. This relationship is crucial for understanding how vector representation and dot products define perpendicularity in geometric contexts.

PREREQUISITES
  • Understanding of vector notation and operations, specifically dot products.
  • Knowledge of linear equations and their geometric interpretations.
  • Familiarity with the concept of slopes and negative reciprocals in geometry.
  • Basic understanding of coordinate systems in a plane.
NEXT STEPS
  • Study vector dot products in detail, focusing on their geometric interpretations.
  • Learn about the properties of slopes and their implications for perpendicular lines.
  • Explore the derivation of line equations from vector forms.
  • Investigate applications of perpendicularity in various mathematical contexts, such as in calculus and physics.
USEFUL FOR

Students of mathematics, educators teaching geometry, and anyone interested in the principles of vector analysis and geometric relationships.

vktsn0303
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The title does not say much. But my actual question is as follows.
Let us suppose a line ax+by=0.
This means A.B = 0 with A and B being vectors, where A = [a,b] and B = [x,y].
Therefore, A is perpendicular to B.

Now my question is if A is also perpendicular to line ax+by=0 (I'm not sure if this makes sense).
I read this statement in a book and I'm not able to understand how this came to be.

Please help me understand this.

Thanks in advance.
 
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vktsn0303 said:
The title does not say much. But my actual question is as follows.
Let us suppose a line ax+by=0.
This means A.B = 0 with A and B being vectors, where A = [a,b] and B = [x,y].
Therefore, A is perpendicular to B.
The equation ax + by = 0 is being interpreted as the dot product of A and B, as you defined them above.
vktsn0303 said:
Now my question is if A is also perpendicular to line ax+by=0 (I'm not sure if this makes sense).
I read this statement in a book and I'm not able to understand how this came to be.
The line ax + by = 0 can be written as y = (-a/b)x, where b ≠ 0. The slope of this line is m = -a/b. The slope of the vector A = <a, b> is b/a. Since -a/b and b/a are negative reciprocals of each other, the vector A and the line ax + by = 0 are perpendicular.
vktsn0303 said:
Please help me understand this.

Thanks in advance.
 
You know the dot product rule If ##\vec{K}=(x_1,y_1)## and ##\vec{L}=(x_2,y_2)##, the dot product will be ##x_1x_2+y_1y_2##.Now Do the same thing A and B.This is the answer why ##\vec{A}.\vec{B}=0##.Lets come to your question.
Do you now write line like a vector? Translate line into a vector form then make a dot product of A.If you do that you will also understand why A is perpandicular to line.
 
Mark44 said:
The equation ax + by = 0 is being interpreted as the dot product of A and B, as you defined them above.
The line ax + by = 0 can be written as y = (-a/b)x, where b ≠ 0. The slope of this line is m = -a/b. The slope of the vector A = <a, b> is b/a. Since -a/b and b/a are negative reciprocals of each other, the vector A and the line ax + by = 0 are perpendicular.

Thanks Mark44, I had forgotten the negative reciprocal rule of slopes for perpendicular lines. Everything makes sense now.
 
Is it OK to say that A is perpendicular to ax+by=0 because A.B=0? Another statement from the same book conveying this meaning.
 
vktsn0303 said:
Is it OK to say that A is perpendicular to ax+by=0 because A.B=0? Another statement from the same book conveying this meaning.
Yes that's true.
 
RyanH42 said:
Yes that's true.

Please explain.
 
vktsn0303 said:
Is it OK to say that A is perpendicular to ax+by=0 because A.B=0? Another statement from the same book conveying this meaning.

I think the above quoted message is misleading here.
My question would have been as follows:
Is it OK to say that A is perpendicular to ax+by=0 because A is perpendicular to B?
 
A is perpendicular to B means A.B=0 you are asking same question again.
 
  • #10
RyanH42 said:
A is perpendicular to B means A.B=0 you are asking same question again.

How is it right to say that A is perpendicular to ax+by=0 just because A is perpendicular to B?
 
  • #11
vktsn0303 said:
Please explain.
If A.B=0, (A=(a,b) and B=(x,y)) we get ax+by=0.Now this means ax=-yb and then y=-ax/b this is a line equation If you write it like a vector you will have to write (x,y) so (x,-ax/b) then If you make dot product rule you get ax+b(-ax/b)=0 so A and line is perpandicular.How can we conclude this idea A.B=0 If you didnt know A.B=0 or you didnt conclude that's not zero you cannot tell me that A is perpandicular to K.Let give an example.A=(3,4) and B=(x;y) now we want to sure that 3x+4y=0 is perpandicular to A.then you need to make 3x+4y=0 like a vector.3x=-4y y=-3x/4 then write it like a vector let's call it K vector.K=(x,-3x/4) now we want to know A.K=? Let's make it , 3x+4(-3x/4)=0 we get zero.So it means A.K=0

If I was made 3x+4y=1 which it means A and B are not perpandicular.Lets see what will happen now.Again we want to know A.K=?. A=(3,4) and K=(x,1-3x/4) If you make dot product you can easily see A.K≠0 means they are not perpandicular.So If A and B are not perpandicular then line and A will be not perpandicular
I hope this helps
 
  • #12
Thank you RyanH42.
 
  • #13
vktsn0303 said:
How is it right to say that A is perpendicular to ax+by=0 just because A is perpendicular to B?
I explained this in post #2.
 
  • #14
Mark44 said:
I explained this in post #2.
Post #2 helped me understand the negative reciprocal rule for perpendicularity.
Post #11 helped me understand the negative reciprocal rule for perpendicularity being applicable, in context of post #10, only after A.B=0 being valid.

The combination of both posts helped me understand everything.
Thanks to both of you.
 

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