Distance between planes | Vectors and spaces | Linear Algebra | Khan Academy
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- Опубликовано: 26 авг 2024
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2010 IIT JEE Paper 1 Problem 51 Distance Between Planes
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Just finished all 144 videos and I wanted to thank you for teaching me. It was definitely interesting, and I'll be looking forward to any future videos.
I am an IIT JEE 2020 aspirant. Your videos have helped me a lot.
Thank You.
Perhaps someone could clarify this... Starting at 11:08 Sal keeps saying "lets find the distance between the two planes" and continues looking for "the distance" for the remainder of the video.
I'm confused because it gives this distance in the initial problem. After reading the questions and comments, everyone else seems equally confused, and none of their explanations are correct either.
Going back to Sal's Normal Vector From plane Equation Video: ruclips.net/video/gw-4wltP5tY/видео.html at 9:18 he explains that the d-value shifts the plane, and doesn't change its tilt.
Isn't this the same as the distance in this situation since the d-value of the blue plane in the equation is 0 in the equation
-x +2y -z=0 (before he multiplies both sides by -1)?
Since the equation of the blue plane in this video is found to be -x +2y -z=0 (before he multiplies both sides by -1), it makes sense that the equation for the red plane can be written as x -2y +z = 6, since it's a distance of 6 units up from the blue plane, however, it contradicts with the given distance of square root (6), which is approx. 2.45.
Since the planes are parallel, doesn't the distance between them remain constant?
What is the actual distance?
I think you'd probably have figured it out, since it's been more than a year but for the ones following this thread, I mention this:
Distance between two planes, Ax + By + Cz = D1 and Ax + By + Cz = D2 is not just |D2- D1| but
|D2- D1|/ sqrt of (A^2 + B^2 +C^2).
Pranav Chaturvedi Thanks for this. I was very confused after his obviously incorrect statements at the end of the video; this makes it pretty much clear. I'm a bit disappointed that he hasn't fixed this... 😔
Finally finished the whole linear algebra playlist. Math is not my major but just because of khan's way of explaining math I fell in love with math. The way you teach I have never learned this way in my whole life. I was just solving questions by remembering formulas, but you changed my whole perspective for math. Now going to learn calculus from your amazing lectures.
Thank you sooooooooooooo much Sal
😍😍😍😭
finally i have completed whole playlist
Thank god
Im just curious but how long did it take you to complete? Im starting out myself
@@Pieruh one a day would be half a year.
Finished 144 videos... MY brain hurts, but I would like to thank you a lot.
It was a long ride but I finally made it to the end! Thanks for these awesome Algebra videos!
If anyone is wondering where to go now that you've finished the playlist, I'd recommend looking up the MIT 18.06 course on linear algebra. Try to do the first two quizzes and if you have any trouble, watch the corresponding MIT lecture videos. Once you feel comfortable with those, finish out the rest of their linear algebra lectures and try the final exam and quiz.
I just got to the symmetric equations in my calculus tonight. I know they're not actual calculus equations but I'm glad to see that they're used in other subjects as well
is it the last video for linear algebra? can't believe i finished the course within 2 weeks.
Can't you just cross (2,3,4) and (3,4,5) without finding random vectors using points like a = 2i + 3j + 4k and b = i + j + k
Wouldn't that get you the normal of the plane as well?
+Will Wei Omg no! Those are the position vectors for points on the plane. Those are NOT a vector on the plane (parallel to the plane). If you cross them, you would get a vector perpendicular to the position vectors, not to the plane itself!
+Will Wei Hold on, I don't fully understand what you're trying to say
no it would get u a random normal that u cant use
Guys guys....(2,3,4) and (3,4,5) are the scaling vectors of the lines. So hes right
@@yonghuiliew8066 yes he is,right
Brilliant
all Linear algebra course done
Just completed the playlist. Thank you for everything Sal!
From the bottom of my heart and my now very tired brain: THANK YOU for this playlist.
when Khan did this, to find vector N * the other part=0, can I use a different vetor(maybe(2,3,4)) to also be =0 and slove it?
Will Wei is right and Arnav Barbaad is wrong. The denominators are the components of the direction vectors for the lines, not the position vectors for a plane. Sal could have just used them instead of working so hard.
Just a suggestion Sal. Could it be at all possible that you include in the description or as links the related videos you talk about. When you mention them in videos they arent always easy to find. Only if its not too much trouble
finished finally
graph theory is next.
SAME QN IN WBJEE 2023 WAS JUST SOLVING IT AND GOT STUCK, 😄
I think he meant the distance between the two closest points should be sqrt(6) at the end of the video
just to clarify: he is solving for ‘d’ in the equation of the first plane, right? because distance is given, but the ‘d’ value of the first is not.
You rock, Sal
To find the distance all you need is a point and the plane equation. You found the point already at the beginning. I don't understand the need to figure out equation of another plane. of course it proved that plane containing other two lines is parallel to the plane in question. Did I miss something?
Sort of, you don't need to find the equation of the second plane, but unless you want to find the answer in terms of A, you have to solve for A before you solve for |d| (which, IMO, should really be |D| as part of the plane equation Ax+By+Cz=D), = |A-1-sqrt(6)*sqrt(A^2+5^2)|
Although, it's easier to use a vector between (any) two unique points with one from each of the lines (e.g. the vector from (1,2,3) on one line to the (2,3,4) on the other = [1,1,1]) and dot it with the normal vector from the plane equation, which must equal 0 since they are orthogonal (or is it perpendicular? I already forgot...), i.e. [1,1,1] dot [A,-2,1] = 0 -> A*1+(-2)*1+1*1 => A = 1
OR
Take that same vector between the two lines, e.g. [1,1,1] and knowing that it is parallel with the defined plane and, therefore, any point on the plane Ax+By+Cz= D also equals Ax_2+By_2+Cz_2, which means Ax+By+Cz= A(x+ the x part of the vector)+B(y+the y part)+C(z+the z part) => Ax -2y + z = A(x+1) -2(y+1) + (z+1) -> 0 = A -2 +1 -> A = 1
Amazing that this is an IIT question. Isn't this US college material?
I dont get it. If they say the distance is square root of 6 then: 1. Why did we have to calculate it ? 2. Why did we get 6 if they say its square root of 6 ? whats the meaning of the difference between the given they provided and the answer we found ?
2nd what daniel ong said. Distance between 2 planes is square root of 6 but the absolute distance is 6??? really confused
The "d" is not "distance it is a constant. Just make it G and you'll understand better
you are awesome :)
All 144 videos? That must only be the videos in some topic, because even four months ago, there were thousands of videos on Khan Academy. Just in case you missed that :) anyway, good going!
144 videos on just linear algebra :)
I dont get it. Distance between 2 planes is square root of 6 but the absolute distance is 6??? really confused
i am still on about vid 20 on this set....hope to get here soon!!!!!!!!!!!!
If both planes are the same equation, don't they share the exact points ?
As you can see at the end, the D value of both equations is different
what if one plane is tilted towards other
If tilted, they will eventually intersect and the distance between them would be zero.
how did (x-2y+z = 0 become x-2y+z = d) and where did A disappear?
There seems to be no audio for me. Is anyone else having the same issue?
nope
horrible explanation.