How To Find The Distance Between a Point and a Plane
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- Опубликовано: 15 дек 2019
- This Calculus 3 video tutorial explains how to find the distance between a point and a plane using the dot product formula and scalar projections of vectors.
Area - Vector Cross Product: • Area of a Parallelogra...
Triple Scalar Product:
• Volume of a Parallelep...
Vector Equations of Lines:
• How To Find The Vector...
The Equation of a Plane:
• How To Find The Equati...
Planar Equation - 3 Points:
• How To Find The Equati...
_______________________________
Lines & Planes - Intersection:
• How To Find The Point ...
Angle Between Two Planes:
• How To Find The Angle ...
Distance Between Point and Plane:
• How To Find The Distan...
Chain Rule - Partial Derivatives:
• Chain Rule With Partia...
Implicit Partial Differentiation:
• Implicit Differentiati...
________________________________
Directional Derivatives:
• How To Find The Direct...
Limits of Multivariable Functions:
• Limits of Multivariabl...
Double Integrals:
• Double Integrals
Local Extrema & Critical Points:
• Local Extrema, Critica...
Absolute Extrema - Max & Min:
• Absolute Maximum and M...
________________________________
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It's really helpful when you go back and explain where the formula come from, rather than just use them. Thank you
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It's very neat how we start by defining a P0 and it then disappears leaving just the components of P1 in the final formula. What a plot twist! Nice video - thanks.
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Professor Organic Chemistry Tutor, thank you for deriving the mathematical formula that is used to Find the Distance Between a Point and a Plane. Deriving formulas in Mathematics increases my knowledge/understanding of the material. This is an error free video/lecture on RUclips TV with the Organic Chemistry Tutor.
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Hey Dude, your explanation is amazing. I want to ask, how to find the closest point from a point to a plane or line. If you've explained it in another video, please let me know. thank you
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Very helpful Sir thanks alot..........btw, what if there are two plane equations, how do you go about solving them?
Alternatively you can think any point on the plane satisfies ax + by + cz = 10 so you can replace the second grouping corresponding to the point on the plane with 10 so you don’t get confused by the sign of d.
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Concretely, I think the reason why ax0+by0+cz0 replace by ax+by+cz is because P0 is in the plane ax+by+cz=-d, thus they are equal to each other.
btw, insightful explanation.
I was taught with a positive d on the right side because it doesn’t matter. You just replace the form ax+by+cz with whatever is on the right and don’t have to worry about 2 sign changes.
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3:14 why did you have to do the projection stuff instead of forming the triangle between points p0, p1, and the point where D touches the plane? And then you can get the trig identity costheta = D/|B| and then rearrange for D= costheta*|B|? What was the purpose of the scalar projection of B onto N?
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Why not just find the line equation (we have direction and position vectors already)? We then substitute x, y and z of the line into the plane equation and solve for the paramater. We then obtain the parameter that we can use to calcualate the point that lies on both plane and the line (which means exactly under the point we are looking for). Then simply calculate the distance between two points and we have the solution.
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I have a lot of questions.. why is n a,b,c isn’t n a normal vector and a,b,c is the equation of the plane ?
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Are you sure D should be -10 and not +10?
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Why are we considering the plane 6x- 3y+2z= 10 to be equal to n vector instead of the horizontal plane perpendicular to point p as depicted in the video?
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What about the shortest distance from a line to a point?
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Why is vector n = (a,b,c) parameters of the equation of the plane? How do you know it?
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What happens if the point is not perpendicular to the plane ? Would you still be able to use the scalar projection of b onto n to find D (the distance)?? Anybody know ???
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How do I find k if the distance k,0 and 2k,0 is 10 units
Damn these videos from "2006" are helping
Please explain how we got d, i got everything else
Must the line the point makes with the plane be normal to the plane?
yes that's the shortest distance
I don't understand why vectors a and b were then compared to vectors n and B, if n is taller then a?
if it is as you just said, the magnitude of n should be equal to the distance b/n the point and the plane, i don't get the logic
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00:02:25 where he says "D is the absolute value of the scalar projection..." he writes "D = comp n B" on the blackboard. I'm not sure what that means exactly or did I misread what he wrote. Anyone have any idea?
So, if you dot B and n, you'll get the projection of B onto n (which is basically what dot product is). But, if you see carefully, the projection B makes on n is equal to the distance D. That's why he wrote D = comp of B on n
comp = component?
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I’m not in calculus but I’m watching this lol
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sorry, why was the denominator the square root of a^2 +b^2 + c^2 please? why is this put in the formula for D, or just from a standard formula?
The distance of a point n with coordinates (a, b, c) from origin is denoted by |n| and is equal to sqrt(a^2 + b^2 + c^2). This is from the equation of distance between two points and substitute the second point as origin i.e (0, 0, 0)
Oh, okay thank you so much, so he substituted the n in, so we have to generally know this square root part or do you know how it was derived, i don't want to confuse but would it be from the a^2+b^2=c^2 formula? Thanks
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actually there is(in my opinion) a simpler way to do it
what is the simpler way?
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I need to find the distance between two planes
find each ones position vector, and then use the distance formula
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