1.

The equation of the plane through the intersection of the planes `vecr.(2hati+6hatj)+12=0` and `vecr.(3hati-hatj+4hatk)=0` and at a unit distance from the origin, isA. `vecr.(2hati+hatj+2hatk)+3=0`B. `vecr.(hati-2hatj+2hatk)+3=0`C. `vecr.(hati-2hatj-2hatk)+3=0`D. `vecr.(2hati+hatj+2hatk)-3=0`

Answer» Correct Answer - A
The equation of the through the intersection of the given planes is
`{vecr.2hati+6hatj+12}+lamda{vecr.3hati-hatj+4hatk}=0`
or `vecr.{(2+3lamda)hati+(6-lamda)hatj+4lamda hatk+12}=0`……………i
It is at a unit distance from the origin
`:.|12/(sqrt((2+3lamda)^(2)+(6-lamda)^(2)+16lamda^(2)))|=1implieslamda=+-2`
Hence the equations of required planes are
`vecr.(2hati+hatj+2hatk)+3=0` and `vecr.(hati-2hatj+2hatk)-3=0`


Discussion

No Comment Found

Related InterviewSolutions