The Equation of a Plane

[@augustusevelyn2702]

..and I finally got it an hour to my finals..*smile* never give up in trying to understand. thanks alot!!

[@fisslewine1222]

Brilliant tutorial; transparent clear and easily understood - so much that I would say there are university lecturers that could do with watching this and gain understanding of how clear and short content can be.

[@Mornys]

Very well structured. I'd recommend this to everyone who's struggling to understand equation of a plane.

[@ZotikosErosElpidius]

Really well explained, and quite thorough. I sincerely thank you for the high quality video. It is much appreciated and truly helped me fully understand the content as opposed to just learning a plug and play formula.

[@BoZhaoengineering]

I am personally digging in the the vector form / parametric form of plane equation. Because from the concept of position vector, basis vector span and linear combination, you finally have a straightforward vector equation(s) to represent the plane. Cool! Thank you Richard!

[@smellymewi]

Dude, you made a complex topic much more simpler for me to understand

[@haidar901]

thank u sir on behave of all the students of linear algebra !!!!

[@muralimohangade9228]

Dear Sir, your explanation about the Equation of a Plane is really excellent. All my doubts are cleared and my problem was solved. I am very much thankful to you. Really Great teaching.

[@dinadayalan9037]

thank you Mr Alrichard, finally I understood something about this topic well. Very clear excellent spoken English and clear explanation.

[@ozzyfromspace]

When you derived the scalar equation of the plane by taking the dot product, I felt like a weight had finally been taken off my mind. Turns out, that was just brain-cells dying for a good cause. Thanks mate!

[@blubberinghumdinger35]

Lucid explanation, absolutely wonderful. Thank you very much :)

[@sebairaf3]

Thats was perfectly explained!

[@DantonCulus]

Awesome how this is well explained, I could even extrapolate to find the intersection point of a segment going through a plane, in a much efficient way than the one I ended to find by myself a month ago...

[@sophiaman2208]

deserves all the likes!

[@nemesiswes426]

Thank you so much, I've been going over this stuff all day and was still not quite sure. I got how to do the equation of a line in 3D already, that was easy but not the equation of a plane and your video just made it seem too easy, lol. I Would have saved myself so much time if I found your video earlier.

[@samdavepollard]

Top quality. Many Thanks for sharing your knowledge.

[@markcavendish7148]

Excellent explanation and colour coding.

[@jjb2723]

found this to be a legitimately well presented and helpful vid... cheers!

[@henrimaurice7395]

This is an amazing video, helped me prep for my Maths A-Level Class tomorrow, many thanks :D

[@sammy5576]

easy to follow, greate content, thanks man

[@habibyella5479]

cheers m8!!! excellent explanation and diagrams. Regards from AUS

[@hohner91]

great vid, really understand all abute this topic!!

[@DiDongi]

Very clear explaination! Thanks sir!

[@AlRichards314]

You can write it either way, but since I chose to write it as 3x - 2y + 6z + D = 0, then when I solved and got D = 2 I shoudl write it as 3x - 2y + 6z + 2 = 0. If I used 3x - 2y + 6z = D and subbed in the point I'd get 3(-4) - 2(1) + 6(2) = D and D = -2 in which case I'd write the equation of the plane as 3x - 2y + 6z = -2. This is equivalent to 3x - 2y + 6z + 2 = 0

[@adamodimattia]

Fantastic stuff, thank you!

[@jorgemercent2995]

Very clear. Thank you.

[@nahidbarghi1707]

Thank You SSo MMuch!!! Grreat Explanation!!

[@AlRichards314]

Yolanda Sangweni, I don't seem to be able to reply to your comment. The calculation is 3(2) - (-2)(6) = 6 + 12 = 18 for the y component.

[@Peter_1986]

Shouldn't this video be listed before the "Lines And Planes" videos in this playlist? It seems like an introductory video for the plane equation. Also, thanks a lot for these videos.

[@AlRichards314]

Thanks for the suggestion. I have since reordered that playlist so it's in a more appropriate order.

[@ntuthukomthiyane5017]

20/10 very clear. thank you sir :)

[@Pho5phoru5]

very helpful video, thank you sir

[@pokepokekoi]

Thank you so much that makes a lot more sense now:)

[@ХристоАврамов-ф3й]

You helped me alot!

[@ozzyfromspace]

Explanation level: god

[@abdirizakmaalim7321]

thank u very much. may God bless u

[@ut2usr2]

Excellent!

[@mosaviadel]

you solved my life's problem ...hips thanks

[@jamesrockford2626]

very nice indeed, thanks

[@diegorivera4417]

And so would the normal be the vector product of v times u?

[@AlRichards314]

Hey. Exactly, so that gives a way to convert from vector or parametric form to the scalar or Cartesian form.

[@mahmoodobada9191]

thank you so much

[@chrischoir3594]

very nice

[@caba8053]

for example 3 how and why was AB and AC obtained. is that not usually multiplication?

[@AlRichards314]

We know the coordinates of point A(-3,4,1) and B(0,2,5). For vector AB, A is the tail and B is the tip of the vector, To get the components of AB we subtract the tail's coordinates from the tip's coordinates. So AB = [0 - -3,2 - 4,5 - 1] = [3,-2,4]. Similarly AC = [3 - -3,6 - 4,-2 - 1] = [6,2,-3]. There's no multiplication when finding components of vectors from coordinates. Maybe you are thinking about the cross-product of two vectors? The reason we need vectors AB and AC is that a plane is defined by a point and 2 non-parallel vectors in the plane. The word "defined" here means that there's only 1 plane in all of 3-Space that has a certain point and 2 specific non-parallel vectors parallel to that plane. So this info specifically describes that particular plane. Also the parametric equations needs the components of these 2 non-parallel vectors in this plane.

[@caba8053]

@@AlRichards314 i really appreciate the help! that small part had been confusing me all night. tail from tip is how i will remember it on the test! thanks a bunch.

[@sukhlearns]

in 8.3, question 8c asks "write a vector eq of a plane that is parallel to the plane 3x+ 5y-z+7=0". And there are no points given. So how can I go about this question?

[@AlRichards314]

Sbrar133 You only need a point if you are looking for a specific plane parallel to 3x + 5y - z + 7 = 0. There are an infinite number of planes parallel to this one (this probably means parallel & distinct as opposed to being the same plane or coincident). Just change the D value to something besides 7, such as 3x + 5y - z - 10 = 0. This has the same normal vector, so it's parallel to 3x + 5y - z + 7 = 0.

[@sukhlearns]

AlRichards314 thank you and your videos are very resourceful. I always watch them before heading to class and they really help :)

[@AlRichards314]

Sbrar133 Hey, thanks for the comment. I'm glad my videos are helpful. That's why I started this channel in the first place.

[@KevinJames1933]

can you explain how the two vectors in the plane are "parallel" to the plane? they do not look like it in the example... thanks

[@AlRichards314]

The 2 vectors parallel to the plane don't have to be parallel to each other. For example drop two pencils or pens on a table. They are resting on the table and so they are parallel to the table (plane), but they likely aren't parallel to each other.

[@KevinJames1933]

okay that makes sense, I wasn't thinking of the plane in 3D.. thanks Al!

[@KevinJames1933]

so this video covers 7.2 and 7.3? (equation of a plane, and properties of planes?)

[@AlRichards314]

Kevin James In the McGraw-Hill text it's sections 8.2 & 8.3 I believe, and yes both sections.

[@ionutincau2225]

Thanks :) this is great

[@HeliosBeats]

this is so good

[@sunshine611531]

I think you mixed up your s and t vectors when writing the parametric equations. you stated that the vector equation was [x,y,z]= [-3,4,5,] + s[6,-5,1] + t[-1,3,2]. but when restating the vector equation you wrote [x,y,z]= [-3,4,5,] + s[-1,3,2] + t[6,-5,]

[@AlRichards314]

Hey. Thanks for the comment. I see what you mean, I switched the vectors in the plane from page 1 to page 2. These are still valid equations, but I will place a note on the video to explain. Thanks again for bringing that to my attention.

[@stephenmuteti2705]

sunshine611531 Thanks......I have also identified the mistake in the working.....was also wondering!!🇰🇪

[@awsomenes01]

+AlRichards314 thank you, these equations can be helped to track a planes directory so it is helping me with what I am trying to make in order to have added safety for planes that is lightweight.

[@Basima9]

thank you!!

[@DaZigstah]

cheers mate

[@sanjaykumar55]

Thankyou sir

[@pwandi3168]

thanks a lot. :)

[@jackflash8756]

That's really weird that the components of the x, y and z in the scalar equation are the coordinates of a normal vector to the plane. Is that just a mathematical coincidence?

[@AlRichards314]

Hey, no that's a property of Scalar Equations. Ax + By + Cz + D = 0 has normal vector [A,B,C]. Cheers.

[@jackflash8756]

Many thanks- really good video .

[@AlRichards314]

@@jackflash8756 Thanks for the comments and I'm happy it helped make sense of plane's equations.

[@IamJacksColon4]

your AP is actually PA..11:33

[@ayonmk]

i love you

[@AthulGanesh]

U da man

[@johnmifsud6814]

Good to hear an American pronounce the letter Z properly - and not "Zeee"

[@alexvarietyhour4919]

wtf!

[@johnmifsud6814]

Good to hear an American pronounce the letter Z properly - and not "Zeee"

[@AlRichards314]

Sorry to disappoint but I'm, Canadian. I guess I'm North American.

[@johnmifsud2698]

@@AlRichards314 All good. I am in Australia, we cant stand American spelling, pronunciation etc. What is worse is the use of British units by Americans - so ridiculous - no one else in the world uses them.

[@AlRichards314]

@@johnmifsud2698 Hey, no problem. I envy you in Australia. I would think your spring/summer is just starting, while our winter creeps closer. Warmest regards.