Why is the gradient of the normal -1/m?

Hey all!
Ive just been getting a bit confused lately as to why the gradient of a normal to a line is -1/m? Why is it not just -m? this is probably really easy but its just been confusing me so much!

Reply 1

username1162972

18

Original post by JakeRStudent

Hey all!
Ive just been getting a bit confused lately as to why the gradient of a normal to a line is -1/m? Why is it not just -m? this is probably really easy but its just been confusing me so much!

Suppose you have two lines in vector format and assume they meet.
Move the meeting point to the origin let one line be column vector a b which gives grad b/a using dot product let the other vector have column vector (x y)
ax+by=0 from product thing then you easily see the relation.

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Reply 2

JakeRStudent

OP

9

Original post by physicsmaths

Suppose you have two lines in vector format and assume they meet.
Move the meeting point to the origin let one line be column vector a b which gives grad b/a using dot product let the other vector have column vector (x y)
ax+by=0 from product thing then you easily see the relation.

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I'm sorry I dont quite follow, could you explain this a bit more thoroughly?

Reply 3

username1162972

18

Original post by JakeRStudent

I'm sorry I dont quite follow, could you explain this a bit more thoroughly?

Do you understand upto the ax+by bit?

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Reply 4

Argylesocksrox

5

Ok, suppose you have a line y=xtanu and a line y=xtan(u+pi/2), these two lines will be perpendicular.

expanding tan(u+pi/2) gives you cosu/(-sinu)=-cotu

hence tanu*-cotu=-1

Reply 5

alkaline.

17

Okay I'm not entirely sure but I feel like what you're looking for is a lot more basic than the answers given

Basically
say m1= gradient of the tangent
And m2= gradient of the normal

M1 x M2 = -1
M2 = -1 / m1

(normal is perpendicular to the tangent)

Reply 6

JakeRStudent

OP

9

Original post by alkaline.

Okay I'm not entirely sure but I feel like what you're looking for is a lot more basic than the answers given

Basically
say m1= gradient of the tangent
And m2= gradient of the normal

M1 x M2 = -1
M2 = -1 / m1

(normal is perpendicular to the tangent)

But why do they equal -1?

And as a note for the above I don'it understand any of that im sorry!

Reply 7

16Characters....

13

Are you familiar with the idea that if we have a line of gradient m then the angle it makes with the positive x-axis (i.e the x-axis moving rightwards) is

\arctan (m)

?

Reply 8

Blazy

8

Original post by JakeRStudent

Hey all!
Ive just been getting a bit confused lately as to why the gradient of a normal to a line is -1/m? Why is it not just -m? this is probably really easy but its just been confusing me so much!

I'm guessing you're looking for a more intuitive answer? Here's a rather dodgy picture. The idea is to rotate the triangle by 90 degrees. Drawing not to scale :smile:

Inspired by Blazy's picture I made my attempt at explaining this.Let you have a look at a detailed explanation in the attached doc.

Original post by Blazy

I'm guessing you're looking for a more intuitive answer? Here's a rather dodgy picture. The idea is to rotate the triangle by 90 degrees. Drawing not to scale :smile:

EXPLANATION ATTEMPT.pdf752.5KB

Reply 10

Matrix123

18

Original post by JakeRStudent

Hey all!
Ive just been getting a bit confused lately as to why the gradient of a normal to a line is -1/m? Why is it not just -m? this is probably really easy but its just been confusing me so much!

The normal is perpendicular to the line. The gradient of the normal is therefore the negative reciprocal of the gradient of the line. If tge gradient of the line is m, the gradient if the normal must be -1/m

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Reply 11

Zacken

22

Original post by Matrix123

The normal is perpendicular to the line. The gradient of the normal is therefore the negative reciprocal of the gradient of the line. If tge gradient of the line is m, the gradient if the normal must be -1/m

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The gradient of the normal is perpendicular to the tangent.

He's asking why does the perpendicularity of the normal and tangent mean that the gradient of the normal is "therefore" the negative reciprocal of the gradient of the tangent. It's quite a good question, actually. :-)

Reply 12

stochasym

12

Then we could talk about Rotation Matrix and the transformation this matrix impose to a vector's coordinates, bearing in mind that we talk about 90 degrees rotation. But this gets more complicated, doesn't it?

Original post by Zacken

The gradient of the normal is perpendicular to the tangent.

He's asking why does the perpendicularity of the normal and tangent mean that the gradient of the normal is "therefore" the negative reciprocal of the gradient of the tangent. It's quite a good question, actually. :-)

Reply 13

nerak99

13

To get a more intuitive idea try this.
Get a piece of quadrille (5mm squared paper) and draw a line from any point to any other.
Lets say from beginning to end is 6 across and 7 up.
From the end draw a line that is 7 across and 6 down.
The two lines will be at right angles.

Try the same trick for a number of examples. Maybe this will help.

Reply 14

Zacken

22

Original post by depymak

Then we could talk about Rotation Matrix and the transformation this matrix impose to a vector's coordinates, bearing in mind that we talk about 90 degrees rotation. But this gets more complicated, doesn't it?

Ooh, that's a nice method as well. I hadn't thought of that. But, yeah - you can always gett more complicated, in fact - once you start working with more than one variable you need to start considering level sets and perpendicularity there is quite insane. :lol:

Reply 15

atsruser

11

Original post by JakeRStudent

Hey all!
Ive just been getting a bit confused lately as to why the gradient of a normal to a line is -1/m? Why is it not just -m? this is probably really easy but its just been confusing me so much!

It's hard to show this without a diagram (I may do one later) but the usual way to demonstrate it is to construct two similar triangles (or even just congruent ..) representing gradients at the point of intersection of the lines. You will find that the gradient of one line is the reciprocal of the other as the rise of the first becomes the run of the other, and vice versa.

The -ve sign comes from the fact that, for one line, as the run increases (i.e. goes in the +ve x direction), the rise decreases (i.e. goes in the -ve y direction).

(edited 8 years ago)

Reply 16

Matrix123

18

Original post by Zacken

The gradient of the normal is perpendicular to the tangent.

He's asking why does the perpendicularity of the normal and tangent mean that the gradient of the normal is "therefore" the negative reciprocal of the gradient of the tangent. It's quite a good question, actually. :-)

Ohhh

thanks for correcting me :five:

Yes, it certainly does sound interesting. Just out of interest, how would you answer that as simply as possible? (Please bear in mind I'm a GCSE student)

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Reply 17

Zacken

22

Original post by Matrix123

Ohhh

thanks for correcting me :five:

Yes, it certainly does sound interesting. Just out of interest, how would you answer that as simply as possible? (Please bear in mind I'm a GCSE student)

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There's a nice picture a few posts up that uses triangles to explain it. Basically, you kind of rotate the triangle and measure the gradient as ratios of sides of triangles and it all works out. The picture should elucidate it all!

Reply 18

Matrix123

18

Original post by Zacken

There's a nice picture a few posts up that uses triangles to explain it. Basically, you kind of rotate the triangle and measure the gradient as ratios of sides of triangles and it all works out. The picture should elucidate it all!