A parallelogram is formed in R^3 by the vectors PA=(3, 2, -3) and PB=(4, 1, 5). The point P=(0, 2, 3)

Find the area of the parallelogram.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

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

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial