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write the equastion the plane through (1,0,0) which is parallel to both of the line r(t)=2t,3t,1-t) and s(t)=(t,1,1-t) can help me with this please
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This is the form of Point- normal plane: From the both lines parallel to the plane, using cross product: -> normal vector n = < -3, 1, -3 > and Point ( 1, 0, 0 ) => Point- Normal plane: -3 ( x - 1) + y - 3 z = 0 Thus Plane equation: -3x + y - 3z + 3 = 0
@linda1 Any comment ?
it is right but is -3x+y-3z=-3
but thank u very muck
calcus 3
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sure thx
how can solve this the plane through the orginwith is perpendicular to both of plane x-y+z=2 and x+y+z=1
Would you open the new post for this question, pls!
sure
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