InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The plane passing through the point (4, -1, 2) and perallel to the lines `(x+2)/(3)=(y-2)/(-1)=(z+1)/(2)` and `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)` also passes through the point |
|
Answer» The general equation of a plane passing throught the point A(4,-1,2) is given by a(x-4) +b(y+1)+c(z-2)=0 This plane will be parallel to each of the given lines only when the normal to the plane is perpendicuale to each of the given lines `therefore (1xxa)+(2xxb)+(3xxc)=0 rarr a+2b+3c=0` `(3xxa)+(-1)xxb+)=0 rarr3a-b+2c=0` on solving (ii) and (iii) by cross multiplication we get `rarr (a)/(7) =(b)/(7)=(c )/(-7) rarr (a)/(1)=(b)/(1)=(c )/(-1)=lambda`(say) `rarr a=lambda , b = lambda` and `c=-lambda` Putting these values of a,b c in (i) we get `lambda (x-4) + lambda(y+1) -lamdba(z-2)=0` `rarr (x-4) +(y+1)-(z-2)=0 rarr x+y -z=1` `Hence the required equation of the plane is x+y-z=1 |
|