Q. 54
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(0, 0, 0), Q(1, −2, 5), R(−1, 2, 11), S(−2, 4, 6)
Step-by-Step Solution
VerifiedPart (a) The four points are coplanar.
Part (b) The quadrilateral PQRS is a parallelogram.
Part (c) The area of quadrilateral PQRS is \(\sqrt{1280}\) 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
\(\overrightarrow{P Q}=\langle 1-(0),-2-0,5-0\rangle\)
\(\overrightarrow{P Q}=\langle 1,-2,5\rangle\)
And
\(\overrightarrow{Q R}=\langle-1-1,2-(-2), 11-5\rangle\)
\(\overrightarrow{Q R}=\langle-2,4,6\rangle\)
And
\(\overrightarrow{R S}=\langle-2-(-1), 4-2,6-11\rangle\)
\(\overrightarrow{R S}=\langle-1,2,-5\rangle\)
Now, let's put all the values in the formula
\(\overrightarrow{P Q} \cdot(\overrightarrow{Q R} \times \overrightarrow{R S})=\left|\begin{array}{ccc}1 & -2 & 5 \\-2 & 4 & 6 \\-1 & 2 & -5\end{array}\right|\)
\(\begin{aligned}\overrightarrow{P Q} \cdot(\overrightarrow{Q R} \times \overrightarrow{R S}) &=1(-20-12)+2(10+6)+5(-4+4) \\&=1(-32)+2(16)+5(0) \\&=-32+32+0 \\&=0\end{aligned}\)
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.
\(\overrightarrow{P Q}=\langle 1-(0),-2-0,5-0\rangle\)
\(\overrightarrow{P Q}=\langle 1,-2,5\rangle\)
And
\(\overrightarrow{Q R}=\langle-1-1,2-(-2), 11-5\rangle\)
\(\overrightarrow{Q R}=\langle-2,4,6\rangle\)
And
\(\overrightarrow{R S}=\langle-2-(-1), 4-2,6-11\rangle\)
\(\overrightarrow{R S}=\langle-1,2,-5\rangle\)
And
\(\overrightarrow{S P}=\langle-2-0,4-0,6-0\rangle\)
\(\overrightarrow{S P}=\langle-2,4,6\rangle\)
The opposite vectors of a quadrilateral are scalar multiples of each other which means opposite sides of a quadrilateral are parallel.
Now, let's find the magnitude of the vectors
\(\underset{PQ}{\rightarrow}=\left<1,-2,5 \right>\)
\(\|\overrightarrow{PQ}\|=\sqrt{1^2+(-2)^2+5^2}\)
\(\|\overrightarrow{PQ}\|=\sqrt{1+4+25}\)
\(\|\overrightarrow{PQ}\|=\sqrt{30} \quad\)
And
\(\underset{QR}{\rightarrow}=\left<2,-4,6 \right>\)
\(\|\overrightarrow{QR}\|=\sqrt{2^2+(-4)^2+6^2}\)
\(\|\overrightarrow{QR}\|=\sqrt{4+16+36}\)
\(\|\overrightarrow{QR}\|=\sqrt{56} \quad\)
And
\(\underset{RS}{\rightarrow}=\left<-1,2,-5 \right>\)
\(\|\overrightarrow{RS}\|=\sqrt{(-1)^2+2^2+(-5)^2}\)
\(\|\overrightarrow{RS}\|=\sqrt{1+4+25}\)
\(\|\overrightarrow{RS}\|=\sqrt{30} \quad\)
And
\(\underset{SP}{\rightarrow}=\left<-2,4,6 \right>\)
\(\|\overrightarrow{SP}\|=\sqrt{(-2)^2+(4)^2+6^2}\)
\(\|\overrightarrow{SP}\|=\sqrt{4+16+36}\)
\(\|\overrightarrow{SP}\|=\sqrt{56} \quad\)
Thus, the opposite sides of quadrilateral PQRS are equal in length. Hence the quadrilateral PQRS is a parallelogram because opposite sides are equal and parallel.
To find the area of quadrilateral PQRS, we will have to find the area of the parallelogram.
\(Area=\|\overrightarrow{PQ} \times \overrightarrow{Q R}\|\)
\(Area=\|\langle 1,-2,5\rangle \times\langle -2,4,6\rangle\|\)
\(Area=\|\langle -32,16,0\rangle\|\)
\(Area=\sqrt{(32)^2+16^2+0}\)
\(Area=\sqrt{1024+256}\)
\(Area=\sqrt{1280}\)
Hence, the area of quadrilateral PQRS is \(\sqrt{1280}\) square units.