Find the distance between skew lines
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- Опубліковано 13 тра 2016
- In this video I define skew lines and go through the process of finding the distance between the two lines using the projection of a random vector connecting the two lines on to the normal vector of the two planes in which the skew lines exist.
best video ever..thank you so much love from suriname
thanks for this! so clarified
thanks so much for this video!!!!!!!!!!!!!!!!!!!!!!!! This is an amazing explanation!! love it!!!!!!!!!!!!!!!!
definitely sharing this link
I realize I'm kinda randomly asking but does anyone know a good website to watch newly released series online ?
It helped me a lot. Thanks
Thank u . Well explained 😃😍
Saved me last minute thanks
why do they have to on parallel planes?
great explanation
NICE EXPLANATION.
how do you know that point (6,3,0) lies exactly at the foot of normal which intersects both the lines??
The normal vector points in the direction perpendicular to both lines. It can be moved anywhere. We have not actually found the point where the lines are closest. We have just found the shortest distance between the two lines. More would have to be done to find where the lines are closest.
thank you so much!
I think it'd be great if you could show why the magnitude of the projection of a onto n is d. Unlike in the case with parallel lines where the distance between the two lines does not change, it's not immediately obvious that in the scenario of skew lines, ||proj_n(a)|| is still d.
I'm not sure how you would explain this to someone who doesn't understand 3d linear transformations, because the way I made intuition of it was a 3d linear transformation that took (i,j,k) as inputs and outputs (n,r,p). Any 2 points you pick on the two lines, the vector formed, a1, will be where n1, r1 and p1 are constants. If you move these two points around, the new vector formed, a2, will be . Notice how only r1 and p1 changes while n1 (||proj_n(a)||) doesn't change, because you're only moving in the direction of r and p, and thus adding multiples of r and p. So it doesn't matter which 2 points you pick, ||proj_n(a)|| remains constant (n1).Now I just pick my 2 points to be the intersection point of n with r and p and you'll find that d = |a| = ||proj_n(a)||. To be clear, I'm no expert in this topic, as I've only learned the concept of linear transformations a few weeks prior, so take what I said with a grain or salt, and feel free to correct me on any errors I've made.
THANKS !
I think you made a mistake at 8:34. When doing the dot product of vectors n and a, you multiplied 7 by -4 for the J component and got -18 when it should've been -28.
yep, I make typos sometimes and since my students are not with me when I make the videos they are not there to yell out that I made a mistake. :) videos are hard.
@@MathBySarah can it really be the shortest distance between the planes too ?
Yes, in the way that I set this up there exist two parallel planes that each contain one of these lines.
The perpendicular distance between the two planes is the shortest distance. So you only need to connect a point from each plain and project it onto their normal vector.
How about if u cross product axn
| Vr X Vp| , that is modulus of the cross product of the two vectors Vr and Vp itself will give the shortest distance.what is the purpose of projecting the distance vector on to the cross product.please clarify
The cross product takes two general vectors in 3-dimensional space, and finds a third vector that is mutually perpendicular to both of them. The shortest distance between two lines is perpendicular to both lines. In 2-d space, this is only possible with parallel lines. We can take two equations for parallel lines, generate a perpendicular line, and solve for where it intersects both of them, and the distance between those two points will tell us how close the two parallel lines are.
Same reasoning applies for 3-D skew lines, except there is only one perpendicular line that makes the closest approach between the two of them. You use the cross product to generate a mutually perpendicular vector to both of the lines. Then make it a unit vector. Then form a vector between any two points on the two given lines (keep it simple by choosing the initial points of the given lines). Take the dot product of this vector and (the unit vector of the cross product vector). That tells you the distance.
NOICE!
I am here 😁
Lol my teacher recommended me this video.😅😅:)
🤟😂
Manoj sir right
Malik ki avaj badli badli si lag rhi hai🤣🤣
@@bhagyeshmahajan4471 😂😂😂
Thank you
Allah bless you ☺️
How did you know that x not for p(s) is the point of which a normal vector of the plane p lies on intersects line r(t)?
I don't know whether your doubt is solved or not but i had a same doubt !!!
if you try to calculate distance by having everything as variables at end you will see whether the fixed point is at the foot of perpendicular or not the distance comes to be constant.
that is same in both case , but the case with the fixed point here (6,3,0) at the foot of perpendicular is easy to visualise though it is not correct to assume it as assumed in the video !! (still answere doesn't change with or without the assumption )
I gained a lot
The answer you got from projecting is a vector, isnt it so that you have to |d|= d * d and then find out the distance? so confused teachers do different
I find a mistake at last step 18 should be 28 and need absolute value
Lol who else came after Manoj Chauhan Sir's post
But the answer i got is -11/√41.
And their was a mistake in your calculation.
U wrote -27+44-18 instead of -27+44-28. Am sorry that I drop this in comment section instead of private message. Thanks ❤
once the two skew lines are contained in parallel planes, then they become PARALLEL.The explanation gets flawed there itself
Think about two planes that are flying in the sky one at 10,000 feet and the other at 20,000 feet. These two planes are flying within parallel planes. But one plane is flying due north while the other plane is flying due east. These two planes will never intersect, and they are not flying parallel to each other, their direction vectors are not the same. Their paths are skew lines and the shortest distance between the two paths that is 10,000 feet.
@@MathBySarah thanks for your reply
@@MathBySarah But one can draw lines on the walls and floor of a room that are skew and those planes (wall, floor) intersect.
@@hooligan1717 you are right, but there are an infinite number of planes in which a line in 3-dimensional space can exist on. If two lines are skew then there is a set of parallel planes in which they exist. That set would then allow us to find the shortest distance between the two lines in the same method. To find these planes they would have to have parallel normal vectors and contain each of the direction vectors. The points on the line would anchor them in space.
Y u stop teaching.?
I am switching careers. I have graded enough homework sets and exams to fill a lifetime, and being a college instructor does not pay enough for me to send my own kids to college so I need to be in a career that will pay a little more than just above the poverty line. I enjoy explaining stuff to my friends and their children, so if you need something just ask, and I might have time to make more videos.
@@MathBySarah
You are just awesome CUZ this video clarified a big doubt I had.
Hope everything is fine :)
English ke chodal hawe ka
why N and vecA start at (6,3,0) thats not true!!!
Line a goes through the point (6,3,0)
A vector valued function does is a collection of vectors whose tail is at the origin and whose head is a point on the line. That being said am not sure what you think is wrong. If you give me a time in the video I can take a look and explain further. I don't know how else to respond to this comment.
@@MathBySarah thanks so much .will get back to you 🙏
Really poor handwriting.
ur account on insta