Why does the product of the gradients of perpendicular lines equal -1 in 3D?

[Yh Hoo]

M1.m2 = -1 ??

Why is the product of the gradients of 2 straight line perpendicular to each other is always equal to -1 ?? Is there any theory to explain this ??

 

[WannabeNewton]

The slopes of perpendicular lines are negative reciprocals of each other. So if y_{1} and y_{2} are perpendicular lines (in standard form) then \frac{\mathrm{d} y_{1}}{\mathrm{d} x} = m and \frac{\mathrm{d} y_{2}}{\mathrm{d} x} = -m^{-1} so their products will simply be -1.

 

[uart]

I think the easiest proof is a geometric one using similar triangles.

- The attachment shows a line, gradient m1, and a perpendicular.

- The two angles marked with dots are equal. (Complementary to common angle).

- By similar triangles d/1 = 1/m1

- By definition (rise/run) the gradient of the perpendicular = -d/1 = -1/m1

 

[sankalpmittal]

Why is the product of the gradients of 2 straight line perpendicular to each other is always equal to -1 ?? Is there any theory to explain this ??

There are many ways to prove this. One and the simpler way is illustrated to you by WannaBeNewton in above post .

Other ways are by trigonometry , use of trigonometrical coordinates , Use of other theorems etc.

I am showing you another simpler way by similarity of triangles ( I hope you know it.)

Let one line with side of course x₁-x and y₁-y be intersected by other line with side of course x₃-x₂ and y₃-y₂ be intersected at angle 90^o

Now you complete their intersection by joining their ends with dotted lines (note one end will be common ie intersection point). You'll notice it will form two right angled triangles .
Now you prove those two triangles similar by their same shape .

By this ,
All corresponding sides of 2 triangles will be proportional , ie
y₁-y/x₁-x = x₃-x₂/y₃-y₂

or₁

m₁=-1/m₂
(Realize one line inclination is > 90 degree hence slope is negative.)

m₁ x m₂ = -1

QED .

 

[willem2]

The slopes of perpendicular lines are negative reciprocals of each other. So if y_{1} and y_{2} are perpendicular lines (in standard form) then \frac{\mathrm{d} y_{1}}{\mathrm{d} x} = m and \frac{\mathrm{d} y_{2}}{\mathrm{d} x} = -m^{-1} so their products will simply be -1.

I think he meant: explain the cause of the result

we have 2 perpendicular lines l and l1. the slope of l = -y/x and the slope of l1 = y1/x1

triangle (O,x1,y1) is similar to triangle (M,x1,x) (two angles are the same)
This triangle is also similar to triangle (O, y, x)

therefore y/x = x1/y1 and if you substitute that in the slope of l1, you'll see it's equal to x/y, so the product of the two slopes = -1

 

[stallionx]

Think in 3D.

Put differently;<m,1> vector is forming 90 degrees with <n,1> vector 2D gradients since original graphs retain perpendicularity.

thus m*n+1=0*sqrt(n**2+1)sqrt(m**2+1)

You can further solve to find
m*n = -1

m=-1/n