InterviewSolution
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InterviewSolution
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Show that the points `(1,-1,3) and (3, 4, 3)` are equidistant from the plane `5x +2y + -7z + 8 = 0.` |
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Answer» Given equation of the plane is `5x+2y-7z+8=0`. Distance of given plane from the point (1, -1, 3) `P_(1)=|(5(1)+2(-1)-7(3)+8)/(sqrt(5^(2)+2^(2)+(-7)^(2)))|` `=|(5-2-21+8)/(sqrt(78))|` `:.P_(1)=|(-10)/(sqrt(78))|=(10)/(sqrt(78))` ...(1) Distance of the plane from the point (3,4,3) `P_(2)=|(5(3)+2(4)+(-7)(3)+8)/(sqrt(5^(2)+2^(2)+(-7)^(2)))|` `=|(15+8-21+8)/(sqrt(78))|` `:.P_(2)=(10)/(sqrt(78))` ....(2) From equation (1) and (2), `P_(1)=P_(2)`. `:.` The given points are equidistant from the plane `5x+2y-7z+8=0`. `" "` Hence Proved. |
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