1.

Find the vector equation of the plane in which the lines `vecr=hati+hatj+lambda(hati+2hatj-hatk)` and `vecr=(hati+hatj)+mu(-hati+hatj-2hatk)` lie.A. `vecr.(hati+hatj+hatk)=0`B. `vecr.(-hati+hatj+hatk)=0`C. `vecr.(-hati+hatj+hatk)=1`D. `vecr.(hati+hatk-hatk)=0`

Answer» Correct Answer - B
The two given lines pass through the point having position vector `veca=hati+hatj` and are parallel to the vector `vecb_(1)=hati+2hatj-hatk` and `vecb_(2)=-hati+hatj-2hatk` respectively. Therefore, the plane containing the given lines also passes through the point with position vector `veca=hati+hatj`. Since the plane contains the lines which aare parallel to the vectors `vecb_(1)` and `vecb_(2)` respectively. Therefore, the plane is normal to the vector `vecn` given by
`vecn=vecb_(1)xxvecb_(2)=|(hati,hatj,hatk),(1,2,-1),(-1,1,-2)|=-3hati+3hatj+3hatk`
Thus, the vector equation of the required planes is
`vecr.vecn=veca.vecn`
`implies vecr.(-3hati+3hatj+3hatk)=(hati+hatj).(-3hati+3hatj+3hatk)`
`impliesvecr.(-hati+hatj+hatk)=0`


Discussion

No Comment Found

Related InterviewSolutions