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Math Help - How do I show that the vector is perpendicular to this line?

  1. #1
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    How do I show that the vector is perpendicular to this line?

    Good Day,

    I've been asked to show that the nonzero vector n = (a, b) is perpendicular to the line ax+by+c=0. I know how to prove this by using gradients but I'd like to show it using vectors.

    I've seen a similar post regarding this but I wasn't able to understand it 'cos the methods weren't clearly shown...

    Assistance is greatly appreciated.
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  2. #2
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    Re: How do I show that the vector is perpendicular to this line?

    What vector represents the line? One way to find it is to use the old-fashioned slope intercept form. y = (-a/b)x - c/b

    This gives such a vector (1,-a/b). The vector product should be scalar zero.
    Last edited by TKHunny; June 29th 2011 at 07:46 PM.
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  3. #3
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    Re: How do I show that the vector is perpendicular to this line?

    Hi. Thanks for the reply.

    However, there's a way to solve this by considering two points on the line, P1 = (x1, y1) and P2 = (x2, y2).

    The vector P1P2 is found as such (x2-x1, y2-y1).

    After this, one finds the dot/ scalar product of the vectors n and P1P2. The result is ax2 - ax1 + by2 - by1.

    What I don't get is how the result of the dot product is 0, therefore proving that the vector n is perpendicular to the line ax + by + c = 0.

    Please do provide detailed steps to help me understand.
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  4. #4
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    Re: How do I show that the vector is perpendicular to this line?

    You didn't really read my post, did you?
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  5. #5
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    Re: How do I show that the vector is perpendicular to this line?

    I did, but I don't see the similarity.
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    Re: How do I show that the vector is perpendicular to this line?

    You have vector (a,b) you need to find vector (k,t) perpendicular to vector (a,b).

    (a,b)*(k,t)=0

    ak+bt=0

    a/b=-t/k

    Hence: t/k=(-a/b) ==> t=-ma, k=mb where m is scalar...

    choose m=1:

    you will get that: vector (k,t)=(b,-a)

    (b,-a)=(-b,a)=(1,-a/b)
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