Nice video, Just have a quick question though: Wouldn't it be faster to just cross the parallel vectors of the lines and then find any point on one of the lines to substitute into the equation? Like if we cross the two parallel vectors we get 3i-6j-3k as a normal vector, and substituting t = 0 into the first line equasion we get the point (6,2,3) Putting it into the plane equation we get: 3(x-6) -6(y-2) -3(z-3) = 0 which simplifies to x-2y-z=-1
You meant taking the cross product of the direction of the intersecting lines instead of parallel vectors, right? Then use a point on one of the lines - yes, that works! If you don't mind I will pin your comment 😁👍
I tried to check both s and t values but they aren't equal one of the axis for example both x and y are ok but z is not or x and z are ok but y is not what should i do with that?
It doesn't matter which of s and t you choose, you just need to make sure you plug t into the line with the parameter t and the s into the line with parameter s. If the two lines do intersect, you should get the same point. So, if you are getting the same x and y but not z, then you may need to check the calculation. Or it's due to the lines do not intersect.
@@johnpaulbalines3498 Good question 👍 When two lines do not intersect, they are either parallel or skew. If they are skew, we can't find the plane; but if they are parallel, we can. Here's the video for it: ruclips.net/video/xO1NDzAyLVs/видео.html
The minus signs come from the plane equation a(x-x0)+b(y-y0)+c(z-z0) = 0, where (x0, y0, z0) is a point in the plane. Sometimes you will see a plus because those numbers in the point can be negative.
Nice video, Just have a quick question though:
Wouldn't it be faster to just cross the parallel vectors of the lines and then find any point on one of the lines to substitute into the equation?
Like if we cross the two parallel vectors we get 3i-6j-3k as a normal vector, and substituting t = 0 into the first line equasion we get the point (6,2,3)
Putting it into the plane equation we get:
3(x-6) -6(y-2) -3(z-3) = 0
which simplifies to x-2y-z=-1
You meant taking the cross product of the direction of the intersecting lines instead of parallel vectors, right? Then use a point on one of the lines - yes, that works! If you don't mind I will pin your comment 😁👍
What a legend. Made it much simpler to understand right off the bat with the explanation of how the plane includes the point of intersection.
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This is a great VIDEO!! I love the aesthetic, the colors, and especially the content! This helped a lot thank you!
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this was great ... clear and simple. Is there to be found a thoroughly explanation of the crossproduct ?
Thank you!! Here's one thorough explanation of cross product www.whitman.edu/mathematics/calculus_late_online/section14.04.html
plus clair que ça! , jamais! Un grand merci pour ce beau travail et cette belle explication.
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this was really helpful
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Very clear and easy to understand, thank you!
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Thank you for the help!
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Thanks a lot, this was very helpful.
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Thank you!
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why are the coefficients of t & s considered to be points on the respective lines, please
Actually, they give the directions of the lines, not the points.
DOOD UNDERRATED CHANNEL
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i love you so much thank you
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I tried to check both s and t values but they aren't equal one of the axis for example both x and y are ok but z is not or x and z are ok but y is not what should i do with that?
Should i choose the s or the t value?
It doesn't matter which of s and t you choose, you just need to make sure you plug t into the line with the parameter t and the s into the line with parameter s. If the two lines do intersect, you should get the same point. So, if you are getting the same x and y but not z, then you may need to check the calculation. Or it's due to the lines do not intersect.
@@GlassofNumbers thank you
@@GlassofNumbers if two lines do not intersect, can we still find the equation of a plane? Hoping for your response.
@@johnpaulbalines3498 Good question 👍 When two lines do not intersect, they are either parallel or skew. If they are skew, we can't find the plane; but if they are parallel, we can. Here's the video for it: ruclips.net/video/xO1NDzAyLVs/видео.html
One question:
in the last step where it says 3( x-1) - 6(y-0) ….
Why is it minus? Or is it always minus? I mean the part in brackets.
Do you mean why we have x minus 1, y minus 0, and z minus 2?
@@GlassofNumbers yes exactly, what determines if it’s minus or plus? It is it always minus as in vector “B - A”
The minus signs come from the plane equation a(x-x0)+b(y-y0)+c(z-z0) = 0, where (x0, y0, z0) is a point in the plane. Sometimes you will see a plus because those numbers in the point can be negative.
@@GlassofNumbers thanks for the reply this clarifies it. Well put together vid as well.