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Sir,
We know that whenever two planes intersect, they intersect in a line.
How to find the equation of that line?
Very urgent please reply soon
Dear Anurag Kishore,
Ans:- Let the two planes be
A1x +B1y +C1z=D1................(1)
A2x+B2y+C2z=D2..................(2)
Now the Direction ratioes normal to the line
(1) are A1 B1 C1...........(3)
(2) are A2 B2 C2.........(4)
now by taking cross product of (3) & (4) we get the vector Q parallel to the required line.
Now the equation of the line is r=P+λQ
where We have got Q. But To find P i.e a point on the line let's consider point a,b,c which lie on the line and hence must satisfy both the equations (1) & (2) So,
A1a+B1b+C1c=D1............(5)
A2a+B2b+C2c=D2............(6)
Now eliminate one variable to make an equation of 2 variables say a & b.
Now by hit and trial method find one pair of a,b satisfying that equation then from it find c also.
So we now have one set of (a,b,c)
So we can find the line.
N.B:- The hit and trial method we have used because there will not be one and only fixed point on the line through which it is passing through.
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.
All the best Anurag Kishore !!!
Regards,
Askiitians ExpertsSoumyajit Das IIT Kharagpur
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