InterviewSolution
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InterviewSolution
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Find the equationfo the plane through points (2,1,0),(3,-2,-2), and (3,1,7). |
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Answer» We know that the equation of a plane passing through three non-collinear points `(x_(1),y_(1),z_(1)), (x_(2),y_(2),z_(2)` and `(x_(3),y_(3),z_(3))` is `|(x-x_(1), y-y_(1),z-z_(1)),(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(x_(3)-x_(1),y_(3)-y_(1),z_(3)-z_(1))|=0` `implies |(x-2,y-1,z-0),(3-2,-2-1,-2-0),(3-2,1-1,7-0)|=0` `implies |(x-2,y-1,z),(1,-3,-2),(1,0,7)|=0` `implies (x-2)(-21+0)-(y-1)(7+2)=z(3)=0` `implies -21x+42-9y+9+3z=0` `implies -21x-9y+3z=-51` `:. 7x+9y-z=17` So, the required equation of plane is `7x+3y-z=17`. |
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