InterviewSolution
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InterviewSolution
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Find the equation of the plane passing through thepoint (-1, 3, 2) and perpendicular to each of the planes`x+2y+3z=5 `and `3x+3y+z=0` |
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Answer» Any plane passing through the point (-1,3,2) is given by `a(x+1)+b(y-3)+c(z-2)=0`………….(i) Now, (i) being perpendicular to each of the planes `x+2y+3z-5=0` and `3x+3y+z=0`, we have, `(a xx 1)+(b xx 2)+(c xx 3)=0 rArr a+2b+3c=0`..........(ii) `(a xx 3) + (b xx 3) + (c xx 1) =0 rArr 3a+3b+c=0`.............(iii) On solving (ii) and (iii) by cross multiplication, we get `a/(2-9)=b/(9-1)=c/(3-6)` `rArr a/-7=b/8 =c/-3 rArr a/7=b/-8=c/3=lambda` (say) `rArr a=7lambda, b=-8lambda` and `c=3lambda`. Putting, `a=7lambda, b=-8lambda` and `c=3lambda` in (i), we get `7lambda(x+1)-8lambda(y-3)+3lambda(z-2)=0` `rArr 7(x+1)-8(y-3)+3(z-2)=0` `rArr 7x-8y+3z+25=0`, which is the required equation of the plane. |
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