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Find equation of the plane through (2, 1...

Find equation of the plane through (2, 1, 4), (1, -1, 2) and (4, -1, 1).

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The correct Answer is:
`(vecr-2hati-hatj - 4hatk) cdot [2hati - 7 hatj + 6 hatk] = 0`
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Find the vector equation of the plane through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x-y+4z=10.

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Find the equation of the plane through the points A(2,1,-1) and B(-1,3,4) and perpendicular to the plane x-2y+4z=10 . Also, show that the plane thus obtained contains the line vecr=(-hati+3hatj+4hatk)+lambda(3hati-2hatj-5hatk) .

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Equation of plane passing through (1,1,0), (-2,2,-1) and (1,2,1) is

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AAKASH INSTITUTE-THREE DIMENSIONAL GEOMETRY -TRY YOURSELF
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