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Find the area of the parallelogram with vertices A=(−1,3,4),B=(0,5,9),C=(1,2,6),and D=(2,4,11).

Question

Find the area of the parallelogram with vertices

A = (−1, 3, 4),
B = (0, 5, 9),
C = (1, 2, 6),
and D = (2, 4, 11).

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Solution

To find the area of a parallelogram given its vertices in 3D space, we can use the formula:

Area = ||AB x AC||

where AB and AC are vectors formed by the vertices A, B, and C, and x denotes the cross product.

Step 1: Find the vectors AB and AC.

AB = B - A = (0,5,9) - (-1,3,4) = (1,2,5) AC = C - A = (1,2,6) - (-1,3,4) = (2,-1,2)

Step 2: Find the cross product of AB and AC.

AB x AC = (AB_yAC_z - AB_zAC_y, AB_zAC_x - AB_xAC_z, AB_xAC_y - AB_yAC_x) = (22 - 5(-1), 52 - 12, 1*(-1) - 2*2) = (4 + 5, 10 - 2, -1 - 4) = (9, 8, -5)

Step 3: Find the magnitude of the cross product.

||AB x AC|| = sqrt((AB x AC)_x^2 + (AB x AC)_y^2 + (AB x AC)_z^2) = sqrt(9^2 + 8^2 + (-5)^2) = sqrt(81 + 64 + 25) = sqrt(170)

So, the area of the parallelogram is sqrt(170) square units.

This problem has been solved

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