Area of parallelogram

Algebra Level 3

Find the area of the parallelogram with vertices A = ( − 2 , 1 , 3 ) , B = ( 0 , 4 , 3 ) , C = ( 4 , 2 , 3 ) , D = ( 2 , − 1 , 3 ) . A=(-2, 1,3), B=(0, 4,3), C=(4, 2,3), D=(2, -1,3).


The answer is 16.

Lawahar Qasid by Lawahar Qasid

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1 solution

Daniel Ferreira
Apr 26, 2015

Fixemos o ponto C, temos então: C D ⃗ = ( − 2 , − 3 , 0 ) \vec{CD} = (- 2, - 3, 0) e C B ⃗ = ( − 4 , 2 , 0 ) \vec{CB} = (- 4, 2, 0) .

Ora, a área do paralelogramo é dada pelo módulo do produto vetorial...

Segue que,

C D ⃗ ∧ C B ⃗ = ∣ i ⃗ j ⃗ k ⃗ − 2 − 3 0 − 4 2 0 ∣ C D ⃗ ∧ C B ⃗ = − 16 k ⃗ C D ⃗ ∧ C B ⃗ = ( 0 , 0 , − 16 ) \\ \vec{CD} \wedge \vec{CB} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ - 2 & - 3 & 0 \\ - 4 & 2 & 0 \end{vmatrix} \\\\\\ \vec{CD} \wedge \vec{CB} = - 16\vec{k} \\\\ \boxed{\vec{CD} \wedge \vec{CB} = (0, 0, - 16)}

Por fim,

∣ C D ⃗ ∧ C B ⃗ ∣ = 0 + 0 + 1 6 2 S ABCD = 16 \\ | \vec{CD} \wedge \vec{CB} | = \sqrt{0 + 0 + 16^2} \\\\ \boxed{\boxed{\boxed{S_{\text{ABCD}} = 16}}}

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