Vector Equation of Line of Intersection of Two Planes
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- Опубликовано: 25 авг 2024
- How do we find a vector equation of line of intersection of two planes x-2y+z=0 and 3x-5y+z=4? We first want to find two points on the line of intersection, and the two points must lie on the plane. After that, we can use the two points to find the direction vector of the line, then use one of the points to write down the parametrization.
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Well explained. Hats off to you😃👏👏
Thank you! Please share this video with others 😁
@@GlassofNumbers Sure thing
i used a random y value to get point B and when i subtracted B-A to get the vector, it did not match to method 2. I used fractions and i am certain i havent miscalculated. What may be the issue then? great video by the way
It doesn't need to match exactly as the one obtained in method 2. The vector that we get should be a multiple of .
Great video, just wanted to say if two lines aren't parallel it doens't mean they must intersect (0:52-1:00)
Thank you! That is true. I didn't say two lines, but two planes at that timestamp period.
I rlly suck at math. Can you tell me how that can be true ?
If they aren't parallel, they can go in different directions and not touch each other at all. E.g. imagine you had two pens, both infinitely long, one in each hand. You could orient them in a way in which they wouldn't intersect. This wouldn't work for planes.@@marthamaywhovier5962
@@marthamaywhovier5962 it is true in 2d, but not in 3d.
if you have two equations that do not cancel out a number, can you multiply one equation by a constant (on both sides) to cancel out a variable?
Ex: eq1 --> -x-3y-4z = 2
eq2 --> 2x + 2y +2z = 5
Yes, cancelling one of the variables is what we try to do. So, that's good!
hello, thank you for your video. I had a question regarding your first approach. I already used your second approach and then attempted to check my answer for V by using the first approach. However, the Z did not cancel out so I had to multiply the equation of the second plane by -2 to get the Z to cancel. Next, I proceeded with your approach and got 2 points and subtracted them from each other. However, the answer did not end up the same as the answer of V with the first approach. Do you have any idea why this is? I could send you a picture of my work so you could check it out, if you would like. Thanks for your time.
The direction vector we get for the line is not unique. The one that you got should be parallel to the one in the other approach. That is, for example, if the direction vector is , then another possible direction vector can be .
Does it matter which one of the planes I subtract ?
No, you can add as well, as long as you can eliminate one of the variables to make it easier for you to find the points
so am i free to pick whichever number ?
Yes, that's correct! Thank you for leaving comments! Please help me share my videos around 😁
hey I have a question so I understand approach two how when you cross them you get a vector perpendicular to the two norm vectors of my plane which result in the direction vector of the line of intersection, however where do I get my point from? since you need a direction vector and a point on the line to create the vector equation. do I just use the norm vector as one of my points
so norm vector 1 was
so my equation would be
+ t
then obviously multiply it out if you want the parametric form
We can get the point by choosing a value for two of the variables and solve for the third from the system of equations.
@@GlassofNumbers so is using the norm vector 1 as position vector correct?
or do we do as you said it, substitute 2 variable and solve for third and get the point?
Can I give a vector equation for the intersection between the two planar equations?
Aren't you referring to this video 🤔
That's the video. Two subtractions get a plane equation. You subtract Z in a special case
@@GlassofNumbers can u do one with three planes pllllss
@@mons9974 I am not sure if this is the one you want to see, it has three planes! ruclips.net/video/NjKmRf0Aa74/видео.html
Didn't much like the video, not enough explanation. Needed approach 2, but everything was super fast forwards, expecially with the n1 (multiplication sign) n2. I mean I understood it because I knew how to do it, but for the people that don't?
Thank you for the feedback! This video assumes knowledge of computing the cross product of two vectors. If one needs to review how to do a cross product, I am happy to make another video just for that. This way, audience who already knows how to do cross product won't be bogged down by calculation.
@GlassofNumbers you're okay, I was just in a bad mood for not understanding this even though I've been trying for hours. Disregard that comment. You're doing great.
just so i can get two point?
Yes, you can choose any two points, as long as they lie in both planes (i.e. they satisfy both equations)
HALWA QUESTION
What question?
@@GlassofNumbers EASY PEASY QUESTION
I WAS WAS SEARCHING FOR IIT ADVANCED WUESTION OF PLANE SOLUTION