Talk:Perpendicular

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Something missing (text error)

The text says "use the SSS congruence theorem for ' and QPB' to " in order to prove perpendicularity, I'm not sure what needs to be before that first tick. — Preceding unsigned comment added by 71.191.226.78 (talk) 15:01, 26 December 2011 (UTC)[reply]

Definition of line perpendicular to plane

The lede to this article previously said:

A line is said to be perpendicular to a plane if 1) the line intersects the plane, 2) the line is not completely contained in the plane, and 3) the line is perpendicular to some line in the plane.

But this is wrong. Any line that intersects a plane is perpendicular to some line in the plane. The correct definition is that a line is perpendicular to a plane if it is perpendicular to every line in the plane. I've changed this definition. —Bkell (talk) 19:16, 25 August 2013 (UTC)[reply]

Er, well, a line is perpendicular to a plane if it is perpendicular to every line in the plane that it intersects. I guess the line won't intersect every line in the plane. —Bkell (talk) 19:22, 25 August 2013 (UTC)[reply]

foot of,perpendicular

What is the foot of the perpendicular if it is not at the bottom as in the diagram? I think I know but it could be made explicit. — Preceding unsigned comment added by 86.145.80.46 (talk) 09:09, 20 July 2017 (UTC)[reply]

Yes, you think correctly, draw an oblique line and construct a vertical line on it ... dare it, make an instance image of it and put it under "Foot of a perpendicular".--Petrus3743 (talk) 12:58, 20 July 2017 (UTC)[reply]

I have just added a verbal description of this term in the section titled "Foot of a perpendicular" so that the definition is not just in a caption. --Bill Cherowitzo (talk) 17:38, 20 July 2017 (UTC)[reply]

Fixing Perpendicular#Graph_of_functions

I am thinking of something along these lines:

${\vec {r}}_{1}=x_{1}{\hat {x}}+y_{1}{\hat {y}}=x_{1}{\hat {x}}+m_{1}x_{1}{\hat {y}}$

${\vec {r}}_{2}=x_{2}{\hat {x}}+y_{2}{\hat {y}}=x_{2}{\hat {x}}+m_{2}x_{2}{\hat {y}}$

${\vec {r}}_{1}\cdot {\vec {r}}_{2}=x_{1}\cdot x_{2}\left(1+m_{1}m_{2})\right)=0$

$m_{1}m_{2}=0$

Guy vandegrift (talk) 23:22, 28 January 2024 (UTC)[reply]

I think the notation might be a bit confusing for the plausible audience for this article (e.g. high school students) without further elaboration and a picture, and could probably be clearer even for an advanced audience; I'd avoid using products xx and yy with differing decorations in favor of using distinct letters.

Ideally this article should be more substantially expanded with a proper explanation of what the dot product is and why a dot product of zero means two vectors are perpendicular, written in as gentle a way as possible. –jacobolus (t) 00:54, 29 January 2024 (UTC)[reply]

Graph

Forgive a newbie, but shouldn't the ninety degree angles have square "markers"? 213.89.13.125 (talk) 18:05, 30 June 2024 (UTC)[reply]