Q. 57
Question
In Exercises 54–57 the coordinates of points P, Q, R, and S are given. (a) Show that the four points are coplanar. (b) Determine whether quadrilateral PQRS is a parallelogram. (c) Find the area of quadrilateral PQRS.
P(3, 4, −2), Q(7, 0, 6), R(2, 1, 7), S(5, −2, 13)
Step-by-Step Solution
VerifiedPart (a) The four points are coplanar.
Part (b) The quadrilateral PQRS is not a parallelogram.
Part (c) The area of quadrilateral PQRS is\(\frac{1}{2}\left ( \sqrt{2336}+\sqrt{1314} \right )\) square units.
A set of points or lines are said to be coplanar if they lie in the same plane.
A quadrilateral is a closed shape and a four-sided polygon that has four edges, four vertices, and four angles.
A parallelogram is defined as a quadrilateral with both pairs of opposite sides parallel and equal.
To show that four points are coplanar, we will use the formula \(\overrightarrow{P Q} \cdot(\overrightarrow{Q R} \times \overrightarrow{R S})=0\)
Let's find the vectors
\(\overrightarrow{P Q}=\langle 7-3,0-4,6-(-2)\rangle\)
\(\overrightarrow{P Q}=\langle 4,(-4),8\rangle\)
And
\(\overrightarrow{Q R}=\langle2-7,1-0,7-6\rangle\)
\(\overrightarrow{Q R}=\langle-5,1,1\rangle\)
And
\(\overrightarrow{R S}=\langle5-2,-2-1,13-7\rangle\)
\(\overrightarrow{R S}=\langle3,-3,6\rangle\)
Now, let's put all the values in the formula
\(\overrightarrow{P Q} \cdot(\overrightarrow{Q R} \times \overrightarrow{R S})=\left|\begin{array}{ccc}4 & -4 & 8 \\-5 & 1 & 1 \\3 & -3 & 6\end{array}\right|\)
\(\begin{aligned}\overrightarrow{P Q} \cdot(\overrightarrow{Q R} \times \overrightarrow{R S}) &=4(6+3)+4(-30-3)+8(15-3) \\&=4(9)+4(-33)+8(12) \\&=36-132+96 \\&=132-132 \\&=0\end{aligned}\)
Hence, the four points are coplanar.
To find that quadrilateral PQRS is a parallelogram. We have to find the vectors of opposite sides because a quadrilateral is said to be a parallelogram if the opposite sides of a quadrilateral are equal and parallel.
\(\overrightarrow{P Q}=\langle 7-3,0-4,6-(-2)\rangle\)
\(\overrightarrow{P Q}=\langle 4,(-4),8\rangle\)
And
\(\overrightarrow{Q R}=\langle2-7,1-0,7-6\rangle\)
\(\overrightarrow{Q R}=\langle-5,1,1\rangle\)
And
\(\overrightarrow{R S}=\langle5-2,-2-1,13-7\rangle\)
\(\overrightarrow{R S}=\langle3,-3,6\rangle\)
And
\(\overrightarrow{SP}=\langle3-5,4-(-2),-2-13\rangle\)
\(\overrightarrow{SP}=\langle-2,6,-15\rangle\)
Thus, opposite vectors of a quadrilateral are not scalar multiples of each other, and opposite sides of a quadrilateral are not parallel.
Hence, the quadrilateral PQRS is not a parallelogram.
To find the area of quadrilateral PQRS, we will have to find the area of two triangles and the sum of those areas.
The formula of the area of the triangle by three points M, N, and O are given by: \(Area=\frac{1}{2}\left\|\underset{MN}{\rightarrow}\times \underset{MO}{\rightarrow} \right\|\).
So, the area of the triangle PQR is:
\(Area=\frac{1}{2}\left\|\underset{PQ}{\rightarrow}\times \underset{QR}{\rightarrow} \right\|\)
\(Area=\frac{1}{2} \left\|\left< 4,-4,8\right>\times \left<-5,1,1 \right> \right\|\)
\(Area=\frac{1}{2} \left\| -12,-44,-16\right\|\)
\(Area=\frac{1}{2}\sqrt{(-12)^2+(-44)^2+(-16)^{2}}\)
\(Area=\frac{1}{2}\sqrt{144+1936+256}\)
\(Area=\frac{1}{2}\sqrt{2336}\)
And the area of the triangle RSP is:
\(Area=\frac{1}{2}\left\|\underset{RS}{\rightarrow}\times \underset{SP}{\rightarrow} \right\|\)
\(Area=\frac{1}{2} \left\|\left< 3,-3,6\right>\times \left<-2,6,-15 \right> \right\|\)
\(Area=\frac{1}{2} \left\| 9,33,12\right\|\)
\(Area=\frac{1}{2}\sqrt{(9)^2+(33)^2+(12)^{2}}\)
\(Area=\frac{1}{2}\sqrt{81+1089+144}\)
\(Area=\frac{1}{2}\sqrt{1314}\)
The area of quadrilateral PQRS is
\(Area=\frac{\sqrt{2336}}{2}+\frac{\sqrt{1314}}{2}\)
\(Area=\frac{1}{2}\left ( \sqrt{2336}+\sqrt{1314} \right )\)
Thus, the area of quadrilateral PQRS is \(\frac{1}{2}\left ( \sqrt{2336}+\sqrt{1314} \right )\) square units.