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Equation of the plane that contains the ...

Equation of the plane that contains the lines `vecr = (hati +hatj) + lambda(hati + 2hatj - hatk)` and `vecr = (hati + hatj) + mu(-hati + hatj-2hatk)` is

A

`vecr(2hati + hatj - 3hatk)=-4`

B

`vecr xx (-hati + hatj + hatk) =vec0`

C

`vecr(-hati + hatj + hatk) =0`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
The lines are parallel to the vectors `vecb_(1) = hati + 2hatj -hatk` and `vecb_(2) = -hati + hatj - 2hatk`.
Therefore, the plane is normal to the vector
`vecn = vecb_(1) xx vecb_(2) = |{:(hati, hatj,hatk),(1,2,-1),(-1,1,-2):}|=-3hati + 3hatj + 3hatk`
The required plane passes through `(hati + hatj)` and is normal to the vector `vecn`. Therefore, its equation is
`vecr.vecn = veca. vecn`
`rArr vecr.(-3hati + 3hatj + 3hatk) = (hati + hatj).(-3hati + 3hatj + 3hatk)`
`rArr vecr.(-3hati + 3hatj + 3hatk) = -3+3`
`rArr vecr.(-hati + hatj +hatk)=0`
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