1.

Equation of line of projection of the line 3x-y+2z-1=0=x+2y-z=2 on the plane 3x+2y+z=0 is

Answer»

`(x+1)/11=(y-1)/-9=(z-1)/-15`
`3x-8y+7z+4=0=3x+2y+z`
`(x+12)/11 = (y+8)/-9 = (z+14)/15`
`(x+12)/11=(y+8)/-9=(z+14)/-15`

Solution :Equation of a plane PASSING through the LINE
`3x-y+2z-1=0=x+2y-z-2` is `3x-y+2z-1+lambda(x+2y-z-2)=0`
`rArr (3+lambda)x+(-1+2lambda)y+(2-lambda)z+(-1-2lambda)`
=0
Since it must be perpendicular to the given plane.
`rArr (3+lambda)3+(-1+2lambda(x+2y-z-2)=0`
`rArr (3+lambda)x+(-1+2lambda)2+(2-lambda)1=0`
`rArr 6lambda-9 rArr lambda=-3//2`
`rArr` Plane is `3x-8y+7z+4=0`
Now line of projection is line of INTERSECTION of PLANES `3x-8y+7z+4=0` and `3x+2y+z=0`


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