Find the equation of the plane through the line of intersection of `vecr cdot(2hati -3hatj-4hatk)=6` and
`vecr cdot (hati -hatj)+4 =0` and perpendicular to vecr cdot(2hati -hatj+ hatk)+8= 0`

Options

A. 47
B. -47
C. 37
D. -37

Answer

A

Explanation

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`.