Find the area of the parallelogram formed by the points $A ( 2 , - 3,5 ) , B ( 7 , - 2,7 )$, $C ( 3 , - 2,11 )$, and $D ( 8 , - 1,13 )$.
A) 816
B) $2 \sqrt { 5 }$
C) 20
D) $4 \sqrt { 51 }$
E) 18

Question

Quizplus · Accepted Answer

The area of a parallelogram formed by vectors $\vec{AB}$ and $\vec{AC}$ can be found using the cross product of these vectors, $\vec{AB} \times \vec{AC}$, and then finding the magnitude of this cross product. First, find vectors $\vec{AB}$ and $\vec{AC}$ by subtracting coordinates: $\vec{AB} = (7-2, -2+3, 7-5) = (5, 1, 2)$ and $\vec{AC} = (3-2, -2+3, 11-5) = (1, 1, 6)$. The cross product $\vec{AB} \times \vec{AC} = (1*6 - 2*1, 2*1 - 5*6, 5*1 - 1*1) = (6 - 2, 2 - 30, 5 - 1) = (4, -28, 4)$. The magnitude of this vector is $\sqrt{4^2 + (-28)^2 + 4^2} = \sqrt{16 + 784 + 16} = \sqrt{816} = 4\sqrt{51}$, which gives the area of the parallelogram.

Find the Area of the Parallelogram Formed by the Points A(2,−3,5),B(7,−2,7)A ( 2 , - 3,5 ) , B ( 7 , - 2,7 )A(2,−3,5),B(7,−2,7)

Find the Area of the Parallelogram Formed by the Points $A ( 2 , - 3,5 ) , B ( 7 , - 2,7 )$