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You need to remember that two planes are parallel if the normal vectors to these planes are parallel.
The problem provides the information that the plane `P_1` passes through the point (4,1,3), hence you may write the scalar equation of plane such that:
`a(x - 4) + b(y - 1) + c(z - 3) = 0`
a,b,c express the coefficients of normal vector `bar n_1` to the plane `P_1` .
You shoud remember that the scalar equation of xy plane is z=0 , hence the normal vector `bar n_2` is `bar k = bar n_2` .
You need to set `bar n_1` equal to `bar n_` 2 such that:
`k(z - 3) = 0 =gt kz - 3k = 0 =gt` kz = 3k => z = 3
Hence, evaluating the scalar equation of the plane passing through the point (4,1,3) and parallel to xy plane yields z = 3.
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