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Find the Cartesianequation of the plane passing through the points `A (0,0,0)`and `b(3,-1,2)`and parallel to the line `(x-4)/1=(y+3)/(-4)=(z+1)/7` |
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Answer» Vector joining point `A` and `B` wil be, `veca = 3hati-hatj+2hatk` Given line is, `(x-4)/1 = (y+3)/(-4) = (z+1)/7` So, `vecb = hati-4hatj+7hatk` is the another vector in the plane. So, a normal to the plane will be, `vec a xx vec b`. Now, `vec a xx vec b = |[hati,hatj,hatk],[3,-1,2],[1,-4,7]|` `=hati(-7+8)-hatj(21-2)+hatk(-12+1)` `=hati-19hatj-11hatk` So, equation of plane will be, `vecr *(vec a xx vecb) = 0` Let `vecr = xhati+yhatj+zhatk` So, required equation will be, `(xhati+yhatj+zhatk)*(hati-19hatj-11hatk) = 0` `=>x-19y-11z = 0`, which is the equation of the plane. |
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