Best Answer:
"Find the area of the triangle determined by the points P(–5, 4, –5), Q(–5, –4, 6) and R(3, –3, 4)"

Another approach (besides using Herron's Formula) is as follows:

The area of a ∆ in 3-space can be found by determining ½ of the magnitude of the 'cross-product' of any two of the three vectors, V₁, V₂, V₃ representing the sides of the 3-D ∆

(I will use the symbol ↗ to denote a vector, if it is not clear from the context)

A(3D∆) = ½ | V₁ X V₂ | ... OR ... = ½ | V₁ X V₃ | ... OR ... = ½ | V₂ X V₃ |,

where 'X' denotes 'cross product'

Any of the three vectors, say for example ↗V₁ = ↗PQ, representing one of the sides of the 3-D ∆, can be determined by the vector subtraction of the two Origin-Vertex vectors, ↗OP – ↗OQ, which can be easily determined by the subtraction of the corresponding pairs of co-ordinates.

V₁ = ↗PQ = ↗OP – ↗OQ = V₁[(–5 –(–5), (4 – (–4), (–5 – 6)] = V₁[0, 8, –11]

V₂ = ↗PR = ↗OP – ↗OR = V₂[(–5 – 3), (4 – (–3), (–5 – 4)] = V₂[–8, 7, –9]

A(∆PQR) = ½ | ↗PQ X ↗PR | = ½ | V₁ X V₂ |

= ½ | [0, 8, –11] X [–8, 7, –9] |

= ½ | [(8)(–9) – 7(–11), 0(–9) –(–8)(–11), 0(7) –(–8)(8)] |

= ½ | [5, –88, 64] |

= ½ √{5² +(–88)² + 64²}

= ½(108.9265...)

= 54.4632...

= 54.5 units² (1 dec pl)

Hope this is helpful! Cheers! ☺

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