Consider the planes `3x-6y-2z=15 and 2x+y-2z=5.` Statement 1:The parametric equations of the line intersection of the given planes are `x=3+14 t ,y=2t ,z=15 tdot` Statement 2: The vector `14 hat i+2 hat j+15 hat k` is parallel to the line of intersection of the given planes.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

D

Explanation

The vectors normals to the given planes are `vecn_(1)=3hati-6hatj-2hatk` and `vecn_(2)=2hati+hatj-2hatk` respectively.
So, a vector parallel to the line of intersection of the given planes is
`vecb=vecn_(1)xxvecn_(2)=|(hati,hatj,hatk),(3,-6,-2),(2,1,-2)|=14hati+2hatj+15hatk`
So, statement -2 is true.
The coordinates of the point where the line of intersection of the given planes intersect with xy-plane i.e. `z=0` are given by
`3x-6y=15` and `2x+y=5`
Solving these two equations we get `x=3,y=-1`
So, the parametric equations of the line of intersection of the given planes is given by `(x-3)/14,(y+1)/3=(z-0)/15=t,tepsilonR`
or `x=3+14t,y=2t-1,z=15t, t epsilon R`.
so statement -1 is not true.