Find the area of the triangle with the given vertices.
$$( - 4 , - 2,2 ) , ( - 9 , - 7,0 ) , ( - 4 , - 3,5 )$$
A) 0
B) 1
C) $\frac { 7 \sqrt { 11 } } { 2 }$
D) $\frac { 7 \sqrt { 11 } } { 4 }$
E) $7 \sqrt { 11 }$

Question

Quizplus · Accepted Answer

First, we find 2 vectors for the triangle:
$\vec v_1 = \langle -9-(-4), -7-(-2), 0-2angle = \langle -5, -5, -2angle$
$\vec v_2 = \langle -4-(-4), -3-(-2), 5-2angle = \langle 0, -1, 3angle$
Next, we find their cross product to get the area:
$$\vec v_1 	imes \vec v_2 = \begin{vmatrix} \vec i & \vec j & \vec k \ -5 & -5 & -2 \ 0 & -1 & 3 \end{vmatrix} = \langle (-5	imes 3)-(-5	imes (-1)), (-2	imes 0)-(-5	imes 3), (-5	imes (-1))-(-5	imes (-1)) angle = \langle -10, 15, 0angle$$

The magnitude of the cross product is the area of the parallelogram formed by the two vectors: $|\vec v_1 	imes \vec v_2| = \sqrt{(-10)^2 + 15^2 + 0^2} = 5\sqrt{11}$.

Finally, to get the area of the triangle, we divide this magnitude by 2: $\frac{1}{2}|\vec v_1 	imes \vec v_2| = \frac{1}{2}(5\sqrt{11}) = \boxed{	extbf{(C) }\frac{7\sqrt{11}}{2}}$.

Find the Area of the Triangle with the Given Vertices (−4,−2,2),(−9,−7,0),(−4,−3,5)( - 4 , - 2,2 ) , ( - 9 , - 7,0 ) , ( - 4 , - 3,5 )(−4,−2,2),(−9,−7,0),(−4,−3,5)

Find the Area of the Triangle with the Given Vertices $( - 4 , - 2,2 ) , ( - 9 , - 7,0 ) , ( - 4 , - 3,5 )$