Distance of a point to a plane | MIT 18.02SC Multivariable Calculus, Fall 2010
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- เผยแพร่เมื่อ 29 มิ.ย. 2024
- Distance of a point to a plane
Instructor: Joel Lewis
View the complete course: ocw.mit.edu/18-02SCF10
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
Oh, will you look at that. I didn't expect an MIT lecture would actually make more sense compared to the textbook I have. Great gob guys!
Most of the time they do make sense since most of the people there are passionate about their subjects and do their best to explain the concepts.
What textbook did you use?
Alternative method: The distance vector between the point P and the plane equals to a coefficient k times the normal vector. Then we can sub the components of the distance vector represented by k plus the coordinate of P into the equation of the plane and solve for k, the coefficient. The last step is to calculate the length of the vector k*normal.
My thank you after seven years since the publication.
Absolutely brilliant!
Having conveniently chosen P as the origin, the coordinates of point Q in the plane will be the normal vector direction multiplied by the determined (normal) distance.
you just unlocked something great
thank you, professor lewis
Amazing explanation..Just wow..
extremely helpful. thank you so much!
Incredible video
太棒了一次看懂。非常感谢您。
这部分知识点看书看了半天也没明白再说什么。看到您说是投影瞬间懂了
Thank you Shlomo
Also, I think it is necessary to give PQ a value in this question because PQ \cdot = 2x+y-2z=4 in this case.
Great video!
Very good explanation.Thanks. May i know , why we are dividing by magnitude of N?
If you are given the plane's equation and a point A,
I find it easier to create a line perpendicular to the plane using the Point (A) given and the normal vector of the plane, which will be the direction vector of this new line.
Then you just have to find the intersection (A') between the Line and the Plane, and calculate the distance between A and A'
I concur.
Thanks very much
I did, pondered for a while, and then figured out I'm out of luck
Good job!👍👍👍👍
im an indian student in class 12..and this belongs to my chapter 11 3D mathematics class 12 CBSE..and its pretty thrilling!
I'm a student in Toronto, Ontario, Canada and we are doing this in grade 12 as well
I'm Indian too and in 12th
@@rofiqel6226 is ur age 16 or 17? coz v r studying this in that particular age mentioned
So what?
@@rofiqel6226So am I. London Ont.
is it the exact same like setting for zero if you are given point q
great vid
thnxx...gr8 video
that is exactly what i said before i click on this video lol. He is a BOSS
Great video, but there is an easier way. We just take the equation E:2x+y-2z-4=0, and normalize the norm vector (that's the Hesse form) by dividing the whole equation by the magnitude of the norm vector. So we get (2x+y-2z-4)/3=0. Now simply plug in the x,y,z coordinates of your point and you're done.
I'm curious, how would you get the x,y,z point of the intersection of "P" and and the plane?
James Rockford The point P and the plane E are not actually intersecting (in that case the distance between the two would be 0)
Jonas Weckschmied
thanks, but hypothetically if a point was above a plane what would be the procedure to find its intersecting point with the plane?
James Rockford So we need a line that goes through point and intersects the plane. Of course that line should be perpendicular (right angle) to the plane, because that is the shortest connection. Now how do we find that line? It's just the normal vector of the plane. In this case it would be (2;1;-2). Or at least, that is the direction vector of the line. The line has to go through our point P. There are lots of different ways of writing an equation for lines, I hope you are familiar with this notation: X = (0;0;0)+s*(2;1-2) where 0,0,0 is point P and 2,1,-2 is our direction vector. So now the point of intersection between the line and the plane can be calculated. Just plug every coordinate of the line equation into the equation of the plane. Remember that the X essentially just means (x,y,z). So 2*(2s)+1*(1s)-2*(-2s)-4=0
Solve for s. Then plug s into the line equation, and you get the x,y,z coordinates of the point of intersection.
This might be a little confusing, but I hope you can follow my thoughts
Note: s is just a random name for the variable I chose, you could ask call it t, r, gamma, or whatever you want
Jonas Weckschmied
thanks appreciate it
what happened to your cos(theta). Isn't the equation vector N*vector PQ = |N||PQ|cos(theta)????
10 months late, but just wanted to clarify to the others that might be watching in the future: he's using vector projection. The one where you use the eqn you mentioned is a different/more complicated method of solving it.
Hi thanks for uploading this video, I just don't understand why you use the coefficients of the x,y and z from the plane equation for the vector for the normal. .??
Thanks so much anyway :)
Vector equation of a plane search it on youtube
Sorry youtube autocorrect
@@adityasahu96 dude delete that comment or correct it
@@abdurrahmanlabib916 shit sorry hahaha
good for you then
Watch out, we've got a badass over here.
You seriously rule.
I thought MIT wld give out of the world level question but pretty easy, JEE MAINS level or boards..... Loved the explanation.....
does it matter if it is vector QP instead of PQ?
yes
oops, never mind. i just figured this out.
I did
who else 16 learning this
I'm 11
@@dark_all_day9311 iam not even born yet!
It's a tutorial he's going to go slow.
This video is kinda old.... Is this method still relevant in modern times?
@Hitogokochi Hahahaha so true
let T(x,y,z) be the "normal point" on the plane to the point in question. Establish any point on the plane. We now have two vectors sharing T as a common endpoint. These two vectors are orthogonal so we know their dot product equals zero. We'll have a system of linear equations for or from which we can solve for T(x,y,z). Knowing the co-ordinates for T(x,y,z) allows us to compute the distance between the off-plane point and T using "3D Pythagoris". Sound reasonable?
You found the joker's day job!
hahaha that's funny. what are you getting angry for
mit >> berkeley
Ahhan, I always thought multivariable calculus was the toughest thing in the world and it turns out we in India learn it in our senior year regular classes... hmm.
There is a lot of things you probably haven't learned in India. Why are you watching this and making an unhelpful comment?
@@nugget9245 if other Indians see this comment, they might relate.
Better change your username to Snowflake
Give a helpful comment. Not a judgmental one.
@@nugget9245 it's not judgmental or whatever. I was just surprised and I expressed that. You're the one being judgmental.
He solved it in 8 minutes and i solved it 4 seconds after seeing the question.
Cuz he's explaining bruhh
I did this in 3 seconds in my head
shut the fuck up