Equation of a Plane Passing Through Two Parallel Lines (2-3t, -t, 1+2t) and (3s, 4+s, 5-2s)
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- Опубликовано: 12 сен 2024
- How do we find an equation of the plane passing through two parallel lines?
The two lines are (2-3t, -t, 1+2t) and (3s, 4+s, 5-2s).
In general, when we want to find an equation of a plane, we need to know a point lying in the plane, and the normal vector of the plane (which is perpendicular to the plane). So, we first need to obtain three points from the two lines (and make sure they are NOT collinear) so that we can find the normal vector!
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Hey man! I am Brazilian and I was having a very hard time trying to discover how to do this... Thankfully I found out your video! Thank you so much!
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God bless you bro far better than my teacher explanation of these
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Thank you this was so great!
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tysm, this was so helpful
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thanks
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Why did u use as t=0,t=1
You can choose other values for t. But the calculation for t=0 and t=1 is easier. 😁
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I got 0 for i,j,k. Why could this be?
You meant for the normal vector of the plane?
This happened to me in my Calc 3 midterm a couple hours ago :). The cross product of two parallel vectors is 0, that might be why.