Question

Find the area of the parallelogram with vertices k(2, 2, 1), l(2, 3, 3), m(6, 9, 3), and n(6, 8, 1).

Answer 1

Let
`\vec K,\vec L,\vec M,\vec N` be vectors pointing the vertices K, L, M, and N, respectively.

The side KL is parallel to and has the same length as the vector

`\vec L - \vec K = (2\,\vec\imath + 3\,\vec\jmath + 3\,\vec k) - (2\,\vec\imath+2\,\vec\jmath+\vec k) = \vec\jmath + 2\,\vec k`

Similarly, the side KN is parallel and as long as

`\vec N - \vec K = (6\,\vec\imath+8\,\vec\jmath+\vec k) - (2\,\vec\imath+2\,\vec\jmath+\vec k) = 4\,\vec\imath+6\,\vec\jmath`

These vectors have magnitudes

`\|\vec L - \vec K\| = √(0^2 + 1^2 + 2^2) = \sqrt5`

`\|\vec N - \vec K\| = √(4^2 + 6^2 + 0^2) = 2√(13)`

and their dot product is

`(\vec L - \vec K) \cdot (\vec N - \vec K) = 0\cdot4 + 1\cdot6+1\cdot0 = 6`

The parallelogram spanned by the vectors
`\vec L-\vec K` and
`\vec N-\vec K` has area equal to the magnitude of their cross product, for which we have the identity

`\bigg\|(\vec L - \vec K) * (\vec N - \vec K)\bigg\| = \|\vec L - \vec K\| \|\vec N - \vec K\| \sin(\theta) \\\\ \implies \text{area} = 2√(65) \, \sin(\theta)`

where
`\theta` is the angle between the sides KL and KN.

From the dot product identity, we have

`(\vec L - \vec K) \cdot (\vec N - \vec K) = \|\vec L - \vec K\| \|\vec N - \vec K\| \cos(\theta) \\\\ \implies 6 = 2√(65) \cos(\theta)`

Then

`\cos(\theta) = \frac3{√(65)} \implies \sin(\theta) = √(1-\cos^2(\theta)) = 2\sqrt{(14)/(65)} \\\\ \implies \text{area} = 2√(65) \cdot 2\sqrt{(14)/(65)} = \boxed{4√(14)}`