# Thread: How do I show that the vector is perpendicular to this line?

1. ## 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.

2. ## 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.

3. ## 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.

4. ## Re: How do I show that the vector is perpendicular to this line?

You didn't really read my post, did you?

5. ## Re: How do I show that the vector is perpendicular to this line?

I did, but I don't see the similarity.

6. ## 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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# vector perpendicular to line

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