1.

If the equation of the plane through the line of interesection of `vecr.(2hati-3hatj+hatk)=1` and `vecr.(hati-hatj)+4=0` and perpendicular to `vecr.(2hati+hatj+hatk)+8=0` is `vecr.(5hati-2hatj-12hatk)=lamda` Then `lamda=`A. 47B. -47C. 37D. -37

Answer» Correct Answer - A
The equation of any plane through the line of intersection of the given planes is
`[vecr.(2hati-3hatj+4hatk)-1]+lamda[vecr.(hati-hatj)+4]=0`
`impliesvecr.[(2+lamda)hati-(3+lamda)hatj+4hatk]=1-4lamda`…………..i
If i is perpendicular to `vecr.(2hati-hatj+hatk)+8=0` then
`[(2+lamda)hati-(3+lamda)hatj+4hatk].(2hati-hatj+hatk)=0`
`implies2(2+lamda)+(3+lamda)+4=0implieslamda=-11//3`
Putting `lamda=-11//3` in i we obtain the equation of the required plane as `vecr.(-5hati+2hatj+12hatk)=47` Hence `lamda=47`.


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