Three vertices of a parallelogram ABCD are A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2). Find the coordinates of the fourth vertex.
VerifiedHint: The diagonals of a parallelogram bisects each other. It means the midpoint of both the diagonals will be the same. If we know the midpoint of one of the diagonals using its vertices then we can find the unknown vertex of the other diagonal if one of them is known.
Complete step-by-step answer:
We have been given the following vertices: A(3, -1, 2), B(1, 2, -4) and C(-1, 1, 2)
Let’s look at the parallelogram first.
Looking to the above figure we can see that
AO = OC, OB = OD
Here O is midpoint at which the diagonals bisect.
So, we need to find the coordinates of O first then we can find the coordinates of D.
From the midpoint formula, we know that
\[O(x,y,z)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2} \right)...........(i)\]
Now we have A and C which are known vertices of the parallelogram so we can use their coordinates to find the midpoint O’s coordinates, here we have,
${{x}_{1}}=3,{{y}_{1}}=-1,{{z}_{1}}=2$ and ${{x}_{2}}=-1,{{y}_{2}}=1,{{z}_{2}}=2$
Substituting the values of respective coordinates from A and C in equation (i) we get
\[\begin{align}
& O(x,y,z)=\left( \dfrac{(3)+(-1)}{2},\dfrac{(-1)+(1)}{2},\dfrac{(2)+(2)}{2} \right) \\
& \Rightarrow O(x,y,z)=\left( \dfrac{3-1}{2},\dfrac{-1+1}{2},\dfrac{2+2}{2} \right) \\
& \Rightarrow O(x,y,z)=\left( \dfrac{2}{2},\dfrac{0}{2},\dfrac{4}{2} \right) \\
& \Rightarrow O(x,y,z)=\left( 1,0,2 \right) \\
\end{align}\]
So, we have the coordinates of O as (1, 0, 2). Now we need to find the coordinates of point D. We will apply equation (i) to the vertices B and D.
Here we have the coordinates of B and O as,
${{x}_{1}}=1,{{y}_{1}}=2,{{z}_{1}}=-4$ and $x=1,y=0,z=2$
Let’s assume the coordinates of D as ${{x}_{2}},{{y}_{2}},{{z}_{2}}$.
Substituting the values of respective coordinates from B, O and D in equation (i) we get
\[\begin{align}
& O(1,0,2)=\left( \dfrac{(1)+({{x}_{2}})}{2},\dfrac{(2)+({{y}_{2}})}{2},\dfrac{(-4)+({{z}_{2}})}{2} \right) \\
& \Rightarrow O(1,0,2)=\left( \dfrac{1+{{x}_{2}}}{2},\dfrac{2+{{y}_{2}}}{2},\dfrac{-4+{{z}_{2}}}{2} \right).............(ii) \\
\end{align}\]
Now equating the x-coordinates, we get
$\Rightarrow 1=\left( \dfrac{1+{{x}_{2}}}{2} \right)$
On cross-multiplication, we get
$\begin{align}
& \Rightarrow 2=1+{{x}_{2}} \\
& \Rightarrow {{x}_{2}}=2-1 \\
& \Rightarrow {{x}_{2}}=1 \\
\end{align}$
Similarly, equating y-coordinates in equation (ii), we get
$\Rightarrow 0=\left( \dfrac{2+{{y}_{2}}}{2} \right)$
On cross-multiplication, we get
$\begin{align}
& \Rightarrow 0=2+{{y}_{2}} \\
& \Rightarrow {{y}_{2}}=0-2 \\
& \Rightarrow {{y}_{2}}=-2 \\
\end{align}$
Similarly equating z-coordinates in equation (ii), we get
$\Rightarrow 2=\left( \dfrac{-4+{{z}_{2}}}{2} \right)$
On cross-multiplication, we get
\[\begin{align}
& \Rightarrow 4=-4+{{z}_{2}} \\
& \Rightarrow {{z}_{2}}=4+4 \\
& \Rightarrow {{z}_{2}}=8 \\
\end{align}\]
So, we have coordinates of D as (1, -2, 8).
Hence, the coordinates of the fourth vertex is (1, -2, 8).
Note: The chances of making mistakes are in cases when the negative sign is ignored during substitution of the coordinates values and if the coordinates of midpoint O is substituted in place of vertex D.
Another approach is finding the midpoint of AC. Then find the equation of BD using the two points equation of line. Then substitute the value of the midpoint of AC as it will lie on line BD, then you will get the coordinate of D. But this will be a lengthy process.
