Find the area of a parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2),
and D(1,10,−5)? Show steps.

Question

Find the area of a parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2), 
and D(1,10,−5)? Show steps.

Theodore Abad · Accepted Answer

#S_("parallelogram") = 15*sqrt(6)~=36.742#

Note that it doesn't matter whether the parallelogram is formed by the vertices A, B, C, D in this order or by the vertices A, B, D. C in this order, any diagonal of the parallelogram divides it in two triangles with equal areas. Then #S_("parallelogram" =2*S_(triangle)#. In this explanation the #triangle ABC# is selected (but any other possible triangle formed with the points A, B. C or D, would do). Repeating the coordinates of the points of #triangle ABC#: A(1,5,0) B(6,10,-3) C(-4,5,-2)

Obtaining sides of the triangle ABC: #AB=sqrt((6-1)^2+(10-5)^2+(-3-0)^2)=sqrt(25+25+9)=sqrt(59)# #AC=sqrt((-4-1)^2+(5-5)^2+(-2-0)^2)=sqrt(25+0+4)=sqrt(29)# #BC=sqrt((-4-6)^2+(5-10)^2+(-2+3)^2)=sqrt(100+25+1)=sqrt(126)#

Using Heron's Formula Triangle ABC (#a=AB, b=AC and c=BC#) #s=(a+b+c)/2=(sqrt(59)+sqrt(29)+sqrt(126))/2~=12.14564# #(s-a)=(-sqrt(59)+sqrt(29)+sqrt(126))/2~=4.46450# #(s-b)=(sqrt(59)-sqrt(29)+sqrt(126))/2~=6.76048# #(s-c)=(sqrt(59)+sqrt(29)-sqrt(126))/2~=0.92067# #S_(triangleABC)=sqrt(s(s-a)(s-b)(s-c))=sqrt(12.141564 xx 4.46450 xx 6.76048 xx 0.92067) = 18.371#

#S_("parallelogram") = 2*S_(triangle ABC) = 2*18.371 = 36.712#

Other way to find the area of the kind of triangle involved in this question is described in: when the triangle is embedded in three-dimensional space

James Atkins · Answer

undefined To find the area of a parallelogram given its vertices, you can use the cross product of two vectors formed by the sides of the parallelogram.

Let's consider vectors $\vec{AB}$ and $\vec{AC}$. 
$$
\vec{AB} = \langle 6-1, 10-5, -3-0 \rangle = \langle 5, 5, -3 \rangle
$$
$$
\vec{AC} = \langle -4-1, 5-5, -2-0 \rangle = \langle -5, 0, -2 \rangle
$$

Now, calculate the cross product of $\vec{AB}$ and $\vec{AC}$:

$$
\vec{AB} \times \vec{AC} = \begin{vmatrix} 
\mathbf{i} & \mathbf{j} & \mathbf{k} \
5 & 5 & -3 \
-5 & 0 & -2 \
\end{vmatrix}
$$
$$
= \mathbf{i} \begin{vmatrix} 5 & -3 \ 0 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 5 & -3 \ -5 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 5 & 5 \ -5 & 0 \end{vmatrix}
$$
$$
= \mathbf{i}(5*(-2) - 5*0) - \mathbf{j}(5*(-2) - (-3)*(-5)) + \mathbf{k}(5*0 - 5*(-5))
$$
$$
= \mathbf{i}(-10) - \mathbf{j}(-10 - 15) + \mathbf{k}(0 + 25)
$$
$$
= -10\mathbf{i} + 25\mathbf{j} + 25\mathbf{k}
$$

The magnitude of this cross product vector is the area of the parallelogram formed by vectors $\vec{AB}$ and $\vec{AC}$.

$$
|\vec{AB} \times \vec{AC}| = \sqrt{(-10)^2 + 25^2 + 25^2} = \sqrt{100 + 625 + 625} = \sqrt{1350} \approx 36.74
$$

So, the area of the parallelogram formed by the given vertices is approximately $36.74$ square units.

Find the area of a parallelogram with vertices A(1,5,0), B(6,10,−3), C(−4,5,−2), and D(1,10,−5)? Show steps.