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Vectors

Cool goIITian

Joined:	23 Jun 2011
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4 Jun 2012 00:13:16 IST

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341

3D geometry

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Vectors

1. Find the equation of the plane passing through the line of intersection of the plane 2x+y-z=3; 5x-3y+4z+9=0and parallel to the line (x-1)/2=(y-3)/4=(z-5)/5

2. Find the equation of the plane through the line (x-1)/2=(y+6)/4=(z+1)/2 and parallel to the line (x+2)/2=(y-1)/-3=(z-7)/5.

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Sagar Saxena

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Joined: 8 Oct 2008

Posts: 8064

10 Aug 2012 21:54:25 IST

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hello

ans.1-->eq. of plane passing through line of intersection OF L1 AND L2 IS GIVEN BY----->

L1+UL2=0 WHERE U IS PARAMETER.

(2X+Y-Z-3)+U(5X-3Y+4Z+9)=0 NOW WE HAVE TO FIND U.

FOR THIS IT IS GIVEN THAT THIS PLANE IS PARAEEL TO VECTOR 2I+4J+5K (FROM LINE)

HENCE THIS VECTOR IS PERPENDICULAR TO NORMAL VECTOR OF PLANE.

[(2+5U)I+(1-3U)J+(4U-1)K].(2I+4J+5K)=0

DO DOT PRODUCT AND GET U FROM HERE.

ANS.2----->THE NORMAL VECTOR OF PLANE IS GIVEN BY

(2I+4J+2K)(2I-3J+5K) =26I-6J-14K (BECAUSE THIS TWO VECTOR LIE IN PLANE)

ALSO POINT (1,-6,-1) LIE ON THE PLANE HENCE EQ. OF PLANE

26(X-1)-6(Y+6)-14(Z+1)=0 SIMPLYFY AND GET ANS.

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