Find the area of the triangle whose vertices are A \left(\right. 1 , 1 , 1 \left.\right) , B \left(\right. 1 , 2 , 3 \left.\right) and C \left(\right. 2 , 3 , 3 \left.\right)

Engineering

Mathematics

Vector Addition and Subtraction

Area of a Triangle

theorem in space

Question

Find the area of the triangle whose vertices are

A (1, 1, 1), B (1, 2, 3)

and

C (2, 3, 3)

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A = (1, 1, 1), B = (1, 2, 3), C = (2, 3, 3)

a

= length of

A B = \sqrt{(1 - 1)^{2} + (2 - 1)^{2} + (3 - 1)^{2}} = \sqrt{5}

b

= length of

B C = \sqrt{(2 - 1)^{2} + (3 - 2)^{2} + (3 - 3)^{2}} = \sqrt{2}

c

= length of

A C = \sqrt{(2 - 1)^{2} + (3 - 1)^{2} + (3 - 1)^{2}} = 3

Semi-perimeter

s = \frac{a + b + c}{2} = \frac{3 + \sqrt{2} + \sqrt{5}}{2}

Area of

Δ A B C = \sqrt{s (s - a) (s - b) (s - c)}

= \sqrt{(\frac{3 + \sqrt{2} + \sqrt{5}}{2}) \times (\frac{\sqrt{2} + \sqrt{5} - 3}{2}) \times (\frac{3 + \sqrt{5} - \sqrt{2}}{2}) \times (\frac{3 + \sqrt{2} - \sqrt{5}}{2})}

= \frac{1}{4} \times \sqrt{(3 + (\sqrt{2} + \sqrt{5})) \times ((\sqrt{2} + \sqrt{5}) - 3) \times [(3 + \sqrt{5} - \sqrt{2}) \times (3 - (\sqrt{5} - \sqrt{2}))]}

= \frac{1}{4} \times \sqrt{((\sqrt{2} + \sqrt{5})^{2} - 3^{2}) (3^{2} - (\sqrt{5} - \sqrt{2})^{2})}

= \frac{1}{4} \times \sqrt{(2 + 5 + 2 \sqrt{10} - 9) \times (9 - (7 - 2 \sqrt{10}))}

= \frac{1}{4} \times \sqrt{(2 \sqrt{10} - 2) \times (2 + 2 \sqrt{10})}

= \frac{1}{4} \sqrt{(2 \sqrt{10})^{2} - 2^{2})}

= \frac{1}{4} \times \sqrt{40 - 4}

= \frac{1}{4} \times \sqrt{36}

= \frac{6}{4} = 1.5 s q . u n i t s