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Find an equation for the plane consisting of all …

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Problem 61 Hard Difficulty

Find an equation for the plane consisting of all points that are equidistant from the points $ (1, 0, -2) $ and $ (3, 4, 0) $.

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Video Transcript

Yeah the question they're asking to find an inauguration for the plane consisting of all points that are equidistant trump. The two points 10 -2 and 340. So in order to find the equation for the plane that is equidistant from these two points we have to first find the values in order to derive the equation. The equation can written in the form of A. In two x minus x zero Plus B into Y -Y0. Let's see into that minors at zero Is equal to zero. Where A. B C. Are components of normal victors of the given plain and Egg 0 Y0 and zero is the point or the coordinates that is lying in the plane. Therefore in order to find the normal victim the point or coordinates first we have to find als midpoint of the given points. So the given points are point A is 10 -2 and point me is 340. Therefore the meet point that is M p me is equal to the some of the coordinates Divided by two. That is MPS equal to one plus 3 x two, zero plus 4 x two And -2-plus 0 x two. So it is equal to two comma two, coma minus one. And so this is equal to the point X zero, Y zero and 00. Mhm. And next year to find them normal victor for the plane. So in order to catholic the normal victim these two points are E and this is me. So coordinator 10 -2 and for B coordinates at 340. So the midpoint is mp that is okay yeah That is 2: -1. Now we're in order to find the normal victim to the plane That isn't it's three D. Form can be presented in a one day form in the line. A. B by the direction vector mp me so therefore M p B is equal to be victor minus mp victor. That is equal to 3 -2. IPs Plus 4 -2, Jacob plus zero. My plus one kick up this is equal to therefore the normal victory is equal to one 21. So we got the point and the normal victim for finding the equation of the plane. So this can be put in the form of the equation that is in two X -60. That is equal to normal victor, one into x minus the point. Let's suppose the point is a 10 -2. The point is The mid .2 to -1 that we are just found out that is equally distant. Did the point b See 2 to -1 and therefore It is equal to one in 2 x -2 Plus two and 2 x -2 plus one in +20 plus one is equal to zero. So this is equal to X -2-plus 2. Why minus four plus zero plus one is equal to zero. This is equal to x plus two, Y plus z minus five equal to zero or X plus two Y plus said equal to five. And this is the required equation of the plane asking the question.

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Calculus: Early Transcendentals

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