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Problem 31 Medium Difficulty

Find an equation of the plane.

The plane through the points $ (0, 1, 1), (1, 0, 1) $, and $ (1, 1, 0) $.

Answer

$x+y+z=2$

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Video Transcript

Hello. So the question is taken from uh equations of lines and planes where we need to find any occasion of playing, forcing 20111018110. So I have done by mistake lee some part of the room. So that will be the creation of plane passing to the point X zero Y 00 can be returning to the plan. X zero Y 00 and abc are normal vector to the plane. Okay so from here if we substitute the value of X zero y zero since it passes 20111019110 these are three equation corresponding to these three points. So if we substitute the value of cesium minus one from two into from one into two we get this value is equal to be okay if we substitute the value of a sequel to be into third we get the value of sees, it is equal to this value. Okay minus saying to explain us one plus Y plus c. And uh if we substitute this value in equation three we get a x minus one plus y minus one minus x minus one plus y plus z is equal to zero. Okay so that it is not equal to eight but it is equal that is not equal to a but it is equal to a because minus two plus one is minus one minus went into minuses. So that will become uh This is not who, this must be one. Okay so from here there will not be any, let me correct this. so and here minus one minus one is minus two. So I know that I tend side we get explains why place they're difficult to do. It is the required equation of plane passing 2.11101 and 110 hope disclosure doubt.