InterviewSolution
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InterviewSolution
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Find the vector equation of the plane through the points (2,1,-1) and (-1,3,4) and perpendicular to the plane `x-2y+4z=10`. |
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Answer» The general equation of a plane passing through the point A(2,1,-1) is given by `a(x-2)+b(y-1)+c(z+1)=0`……………(i) If the point B(-1,3,4) lies on plane (i), then we have `a(-1-2)+b(3-1)+c(4+1)=0` `rArr -3a+2b+5c=0`………….(ii) If the plane (i) is perpendicular to the plane `x-2y+4z=10`, then we have `(1 xx a) -(2 xx b)+(4 xx c)=0` `rArr a-2b+4c=0`.............(iii) On solving (ii) and (iii) by cross multiplication, we have `a/(8+10)=b/(5+12)=c/(6-2)` `rArr a/18=b/17=c/4= lambda` (say) `rArr a=18lambda, b=17lambda` and `c=4lambda` Substituting these values of a,b and c in (i), we get `18lambda(x-2)+17lambda(y-1)+4lambda(z+1)=0` `rArr 18(x-2)+17lambda(y-1)+4lambda(z+1)=0` `rArr 18(x-2)+17(y-1)+4(z+1)=0` `rArr 18x+17y+4z=49`. Required equation of the plane in vector form is given by `vecr.(18hati+17hatj+4hatk)=49`. |
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