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!!
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.
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.
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.
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.
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?
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
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?
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.
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!
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.)
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.
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
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).
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 ...................
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.
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 TH-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. ;)
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.
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.
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.
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
@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.
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
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.
How to be sure that the intersection line is not parallel to the x-axis?
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 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.
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.
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?
I havent taken Linear Algebra yet but listening how it was approached in a similar but different way really helped me understand the concept
superb explanation. you make it very easy to visualize and the way you explain it helps to get the intuition. thankyou sir.
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
Fantastic video- this is better than any other video/lecture I have found on this topic.
Superb lecture
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?
Thank you explain this so much easier than book. Vectors are much more easy now
This concept become crystal clear after watching your video.Thanks a lot.
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.
where was this all my life
which software did you use to write it this way ?
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!
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.)
Thank you, Scott Lawson. You are the bomb.
Awesome brother ....
is these a way to get the plain equation based on the vertices in each corner?
very interesting take on how to solve these kinds of problems, thank you!
what if it is not directly double . like 5 and 2? what do you do? do you still double ?
Thanks sir..👍❤️
Very helpful! Thanks!
Excellent explanation!
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.
how to determine which one need to be 0?
Thank you I finally got my WebWork question correct!
Good explanation, I was satisfied with the explanation
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
Isn't 15 -> -15 because the y in cross product is multiplied by -1?
Ai - Bj + Ck ?
than you, you helped me visualise the shiet out of these planes
we can put y=0 or z=0 instead of putting x=0 for same answer. Am I right?
i believe so
is it me or was there a problem with the cross product at 6+9
i got -6 - 9 = -15
how is the direction of the line determined?
Very well explained. Thank you so much!
shouldnt the 23 be a -7 becuase -5X3 = -15 on the last mutliplied set for the crossproduct
W teacher!
good job man
thank you man
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).
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?
would be great if you didnt have one that was a multiple of another how the fuck can a guy solve these other wise.
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...
Nice, thanks!
this is absolute fukin gold
thank you so much! this was so helpful!
thanks man
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.
dude you are awesome!
Idk why most people solve this by equating x u and z terms... This is actually how you imagine and do it
thanks! well done.
was having trouble visualizing it until 3:20. Just what i needed.
zed sounds good enough..😃
yh I think u get the same line but translated elsewhere.
With this, I am going to avoid an F in a test. Thanks!
Thank u
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.
Quite helpful
Thanks man!
thank you ,sir !
thank you so much!
Thank you :D
Thx a lot mann
Ah I see.. thank you!
This was great but you forgot to turn your dog whistle off
no man it shouldnt be so.. he has done it correctly...coz the general equation is (answer1-answer2) not addition of both.. :)
Oh God, u should return to TH-cam
*zed*
's dead. zed's dead.
what if none of the planes contain x=0
Pls help me out
Find the equation of the line of intersection of the planes
13x+13y-6z=8
2x+2y-6z=22
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 TH-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
the audio buzz is killing me
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.