Finding the vector equation for a line that intersects two planes - Linear Algebra -
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- Опубліковано 19 січ 2012
- How to find the line where two planes intersect or meet. Using the cross product and centering the plane on some point, we can put the equation in parametric form. In this video I demonstrate one method for calculating the cross product. • Very easy way to calcu...
You literally taught me more in 11 minutes (~5 min @ 2x speed) than my professor did in two 50-minute sessions. Midterm in 3 hours, pretty sure this raised my grade at least 5%. Thank you!!
although 9 years has passed, you still saved my life before a midterm
You shouldn't assume that if you set x=0 and solve for that, that you'll get the right answer in the end. Instead, SOLVE the equation-system in terms of either x,y or z. then give your answer in vectorform or whatever form they ask for.
Only given the low values for the normalvektors gives you a significant chance of finding a correct positionvector if you set x=0, nothing else. As said, solve the system.
How are you supposed to solve the two equations with three unknowns? Or are you setting x to 1?
The line should definitely cross X=0 unless it is parallel to the x axis. So this method should be fine
@@FBR2169 do you mean perpendicular to the x-axis?
@@mosabqaissiah5198 yes that XD
Frans Rewl the point has to lie on the line of intersection and x=0 may not lie on that line
Excellent video and excellent explanation. I have watched many videos across youtube and I prefer your style (just like the style of KhanAcademy) the best.
@9:07 you turn z = y(13/9) and y = z(9/13) into y= -18 and z = -26. without showing how you got there I am completely confused.
How to be sure that the intersection line is not parallel to the x-axis?
Excellent explanation!
Thanks for such a valuable explanation. I have a suggestion about finding the y and z components of the common point. If you sum the two equations up 3z's will automatically cancel and you easily get y=-18.
I was just thinking that
Indians will always do that.
Thank you, Scott Lawson. You are the bomb.
Very helpful! Thanks!
Thank you explain this so much easier than book. Vectors are much more easy now
also, what would you do if there was no way to set the equation for a equal to equation for b?, for instance, what if equation b were equal to 7, not 6, multiplying that equation by two and setting it equal to equation a would not work, correct? is there a general way of going about this so that it always works?
superb explanation. you make it very easy to visualize and the way you explain it helps to get the intuition. thankyou sir.
7:00 why do that ? when you multiplied the numbers, you can't solve y and z because they don't eliminate. Right now its -3z and +3z in each equation. Use elimination (so substitution) and add those equations for x=0. You can find a precise point of the line when you make x=0, add line A and B, the 3z cancels, and youre left with y=-16. Then put y=-16 into either equation, you're left with z=-26, so now your vector is x,y,z=(0,-16,-26) (a starting point but any starting point would do) + t(3,15,23) (your position vector from n1xn2), or written parametrically. x=0+3t, y=-16+15t, z=-26+23t. When you doubled the number, you have not eliminated y/z from the equation...the number is irrelevant.
Awesome brother ....
I havent taken Linear Algebra yet but listening how it was approached in a similar but different way really helped me understand the concept
how come one of the the normal vector is on the plane? the blue one, i thought that normal vector must be perpendicular to the plane into three dimensional space? thank you ! great lecture!
Fantastic video- this is better than any other video/lecture I have found on this topic.
Superb lecture
Very well explained. Thank you so much!
Thanks for the help, but I do have one question. One of the questions in my textbook about this is set the same way with two standard form equations, but one of them is set equal to 0. Your method is about if the two equations are set to real numbers, pretty much anything but the infinities and 0. So, what should I do?
9 years later and around 7:40ish I threw up my hands and said "THANK YOU" lmao , I needed this so badly thank you so much !!
Thanks sir..👍❤️
This concept become crystal clear after watching your video.Thanks a lot.
Hey, I believe what your mistake is, you are to take the negative of the second variable... if you see how you find the determinant it is by (-1)^(row number + column number)*det(matrix), and the thus it becomes -(-6-9) = 15... this is how I was taught the cross product :D
very interesting take on how to solve these kinds of problems, thank you!
Something that feels weird when doing this is that the equation for a plane and the equation for a line are in completely different forms. One is an implicit equation constraining three variables, the other is an explicit function of 1 input. The plane equivalent of the equation of a line given feels like it should be a point and two vectors. (Of course I may have an unorthodox perspective after seeing the line intersection between two planes gotten by a single multiplication of the planes in question.)
Good explanation, I was satisfied with the explanation
good job man
where was this all my life
which software did you use to write it this way ?
Nice, thanks!
thank you so much! this was so helpful!
thank you man
im confused...first of all when you were working the equation at 8:11 how are you just able to ignore the 12?? , second, when i plugged in your final value for y (9/13*z) into the equation for the vector a instead of vector b as you did, i got z = -2, which is completely different. How do i know which equation to plug these values into?
If you hold x(or y,z, as long as it is any one of the unknowns) a different constant, you’ll get a different vector perpendicular to two normal vectors. However, the right vector should always be linearly relevant, not necessarily identical.
thanks! well done.
was having trouble visualizing it until 3:20. Just what i needed.
how to determine which one need to be 0?
What were you doing during 8 min ~9 min over an equation group? If you hold x=0, you’ll find 3z and -3z can offset each other and find y=y=-18 in no more ten seconds. Your twisting calculation made me like a scratch cat.
is these a way to get the plain equation based on the vertices in each corner?
thanks man
is it me or was there a problem with the cross product at 6+9
i got -6 - 9 = -15
dude you are awesome!
than you, you helped me visualise the shiet out of these planes
Thank u
Show me how to find the line of intersection of the plane containing (0,0,1),(1,1,2) and (2,2,3) with the plane containing the points (1,2,3), (0,1,2) and (-2,-1,1).
Thanks man!
thank you ,sir !
this is absolute fukin gold
thank you so much!
what if it is not directly double . like 5 and 2? what do you do? do you still double ?
yh I think u get the same line but translated elsewhere.
Thx a lot mann
Thank you I finally got my WebWork question correct!
Thank you :D
Isn't 15 -> -15 because the y in cross product is multiplied by -1?
Ai - Bj + Ck ?
we can put y=0 or z=0 instead of putting x=0 for same answer. Am I right?
i believe so
What we are looking for is a point that belongs to both of the planes, if the point fits in both equations of the plane then it definitely belongs to each of them. That is why we are sure that it is on the line of intersection. In the same manner we could set x=2,6 and solve for it.
It can happen that the line does not equal 0, or 2.6 or anything else we want to set it, but in that case in the equation of the plane we won't have a*x. we would have a fixed value. e.g 5+by+cz=7
zed sounds good enough..😃
can we say that at SOME POINT, both the x's will be 1 or any other number, instead of 0 and calculate it from there?
shouldnt the 23 be a -7 becuase -5X3 = -15 on the last mutliplied set for the crossproduct
You can't just assume that x=0, i've tried it with two planes and the values that i get don't satisfy both planes.
I don't like the mixing up of co-ordinate notation with vector notation. A co-ordinate is not a vector, and vice versa. t(a, b, c) is a little meaningless when you think about it.
we can perform cross product of two vector only when two vector lie int he same plane , so sir u have done the cross product of two normal which lie in the different plane so how that cross product is possible ...................
These 2 vectors can be moved into one plane...
Quite helpful
would be great if you didnt have one that was a multiple of another how the fuck can a guy solve these other wise.
how is the direction of the line determined?
Idk why most people solve this by equating x u and z terms... This is actually how you imagine and do it
Ah I see.. thank you!
With this, I am going to avoid an F in a test. Thanks!
*zed*
's dead. zed's dead.
no man it shouldnt be so.. he has done it correctly...coz the general equation is (answer1-answer2) not addition of both.. :)
Pls help me out
Find the equation of the line of intersection of the planes
13x+13y-6z=8
2x+2y-6z=22
what if none of the planes contain x=0
Oh God, u should return to UA-cam
This was great but you forgot to turn your dog whistle off
why can't you just say z? ....zed... this guy!!
+Shawn Huetter He's British you dumb ass
Jack Lloyd Smart of you to call me a dumb ass for something that has no coalition with my intelligence(may show how much diversity I’m lacking on other countries pronunciation of certain words and letters.) I am, however, sorry you got butt hurt by such a comment on UA-cam. But to set the facts slightly straight... it’s not just the Brits that say “zed.” Thank you though, for helping expand my knowledge, this is something I didn't know prior to your childish comment. So maybe this will help you rethink the way you reply to someone before you make yourself look like a spoiled 8 year old. ;)
guys the australians,Indians and NZ guys say zed too I guess
+Jack Lloyd Wait, he's British??? He sounds American. Maybe he doesn't live in the US, but no one just changes the way they pronounce things
Sorry, I'm Canadian
The cross product is not correct. The first column should be 2, -5, -3. That makes the X-product .
Why do you say that the first column be [2, -5, -3] instead of [2, -5, 3]? The equation is 2x - 5y + 3z
Do people really have issues visualizing planes?
Coming from someone who is just starting his third semester of calculus and had only ever dealt with 2D coordinate systems, the answer is an emphatic yes. I've been dealing with lines and single-variable functions for so long I'm used to being able to simply look at the equation for a line and instantly visualizing it in my head. While I can visualize what a generic plane might look like, there is still a significant gap between seeing the equation for one and turning the numbers into a mental picture.
+TheRestrictedgamer Literally in the same boat right now! hope you passed man. I just started calc three and the visulizations are kicking my ass
+Mitchell “Mottz” Allen I guess playing a lot of video games i've never had an issue visualising space.
bruhhh, yes im 17 and am in my last year of high school, i dont really have trouble visualising a plane but I never visualised that the normal to both the normals of the plane is parallel to the line lol. idk i just cant think like that for some reason.
Yes nerd
Sir you have very bad habits, you are really making this a lot harder due to those bad habits, your Matrix row and column is completely wrong then you speak about 0*3 which equals 0 that you don't even completely finish what you were saying going completely off topic, to continuing on as nothing had ever been said.
the audio buzz is killing me