Area of parallelogram

Question

Find the area of a parallelogram PQRS where P = (2,6,4), Q = (−5,−3,1), R = (−3,1,1), and S = (−10,−8,−2)

Answer 1

One thing to notice if you plot the vertices, than naming the vertices in order for forming a parallelogram results in PQSR.

To find the area of that parallelogram we want to determine the magnitude of the cross product of two of the vectors that are two sides of the parallelogram starting with the same vertex.

Let's use vector PQ and vector PR.

Vector PR starts at (2,6,4) and ends at (-5,-3,1), so in standard position <-7, -9, -3>.

Vector PR starts at (2,6,4) and ends at (-3,1,1), so in standard position <-5, -5, -3>.

The cross product of those two vectors (using diagonals instead of expansion by minors):

i j k i j

-7 -9 -3 -7 -9

-5 -5 -3 -5 -5

results in:

27i +15j +35k - (21j + 15i +45k) = 12i - 6j - 10k

The magnitude of the cross product:

√[12² + (-6)² + (-10)²] = √280 = 2√70 ≈ 16.73 sq units

This Desmos graph pictures the parallelogram and also shows that using vectors SQ and SR gives the same result:

desmos.com/3d/itryaggoq4