One side of the parallelogram is the vector PA and another is PB. Let the fourth vertex be C.

The vertex A is (0, 2, 3) + (3, 2, -3) = (3, 4, 0)

The vertex B is (0, 2, 3) + (4, 1, 5) = (4, 3, 8)

The...

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One side of the parallelogram is the vector PA and another is PB. Let the fourth vertex be C.

The vertex A is (0, 2, 3) + (3, 2, -3) = (3, 4, 0)

The vertex B is (0, 2, 3) + (4, 1, 5) = (4, 3, 8)

The fourth vertex C is A + PB = (7, 5, 5)

Now, the diagonals of the parallelogram are:

AB = (3, 4, 0) - (4, 3, 8) = (-1, 1, -8). |AB| = sqrt (1+1+64) = sqrt(66)

PC = (7, 5, 5) - (0, 2, 3) = (7, 3, 2), |PQ| = sqrt(49+9+4) = sqrt(62)

Let the angle APB be equal to x, this gives:

|AB|^2 = |PA|^2 + |PB|^2 - 2 |PA| |PB| cos x

=> 66 = 22 + 42 - 2*sqrt(22*42)*cos x

=> cos x = 1/sqrt(22*42)

=> sin x = sqrt ( 1 - 1/(sqrt 22*42)^2 )

=> sin x = sqrt (22*42 - 1)/sqrt(22*42)

Taking an altitude from vertex A for the triangle PAB, we have

h = |PA| sin x = sqrt(22) sqrt (22*42-1) / sqrt(22*42)

The area of the triangle is (1/2)*|PB|*h. The parallelogram is twice this area or the required area is:

2*(1/2)*|PB|*h

=> sqrt(42)*sqrt(22)*sqrt (22*42 - 1)/sqrt(22*42)

=> sqrt(22*42 - 1)

=> sqrt(923)

=>30.38

**The required area of the parallelogram is 30.38 **