Find area of parallelogram

Question

Verify that the points are the vertices of a parallelogram and then find its area. (2,-1,1),(5,1,4),(0,1,1),(3,3,4).

Answer 1

Treating the points as position vectors in R³ we just need to show that two sides are given by the same vector.

 

⟨5,1,4⟩ - ⟨2,-1,1⟩ = ⟨3,2,3⟩

⟨3,3,4⟩ - ⟨0,1,1⟩ = ⟨3,2,3⟩

 

That is one pair of parallel sides. Both are the same vector, ⟨3,2,3⟩, so we have a parallelogram.

However, we need the vector for the other pair of parallel sides to find the area.

 

⟨2,-1,1⟩ - ⟨0,1,1⟩ = ⟨2,-2,0⟩
⟨5,1,4⟩ - ⟨3,3,4⟩ = ⟨2,-2,0⟩

 

The other pair is given by vector ⟨2,-2,0⟩.

 

The cross product is a vector whose length is the parallelogram's area.

 

⟨3,2,3⟩ × ⟨2,-2,0⟩ =

 

| i   j  k |

| 3  2  3 |

| 2 -2  0 |

 

= [(2)(0) - (3)(-2)]i + [(3)(2) - (3)(0)]j + [(3)(-2) - (2)(2)]k

 

= 6i + 6j - 10k

 

= ⟨6,6,-10⟩

 

Area = √[6² + 6² + (-10)²] = √172 = 2√43

Answer 2

You want to do this in three steps.

 

(a) show all four points are coplanar

 

(b) show opposite sides are parallel

 

(c) find the area (find lengths first)

 

 

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