Vectors - Equations of plane Watch

davidsmith
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Report Thread starter 10 years ago
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If a plane has the following equation

r= (3,3,5) + t(-3,-1,2.30) + s(-2,-2,-3)

and Point S has coordinates of (4,1,-1)

what is the shortest distance between them?!
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Cities
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I remember there is a standard method for doing this; if you have the Heinemann FP3 book, it's definitely outlined in there. Unfortunately I can't remember that, so I'll guess, but check with someone else / book. The plane is made up of two vector components, so to find the vector which is perpendicular to them both, use the vector product. Then find the equation of this perpendicular vector which goes through the point S. The distance between the point S and the point at which the line intersects the plane is the shortest distance.
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Operator
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n1r4v: You're right, but you've forgotten to mention:

1. You need to find the equation of the line whose direction vector is a scalar multiple of the vector normal to the plane and whose initial vector is S. It'll be of the form {\bf r} = {\bf a} + \lambda{\bf b}.

2. Find the equation of the plane in terms of {\bf r}\cdot{\bf n} = p. But substitute {\bf r} = {\bf a} + \lambda{\bf b} into the plane equation to find the value of \lambda for which our line intersects the plane.

3. Using this value of \lambda, find the coordinates of the point for which the line and plane meet. The shortest distance between this point and S can then be found by Pythagoras.

It's not an easy question (for me at least), so if you have any trouble getting your head round it, just ask.
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