# Find a nonzero vector orthogonal to the plane through the points P, Q, and R, and area of the triangle PQR.

Take note of the following points:
$P(1,0,1) , Q(-2,1,4) , R(7,2,7)$

• Find a nonzero vector orthogonal to the plane through the points $P, Q$, and $R$.
• Find the area of the triangle $PQR$.

The purpose of this question is to find an orthogonal vector and the area of a triangle using the vectors $P, Q,$ and $R$.

A vector is essentially any mathematical quantity that has a magnitude, is defined in a specific direction, and the addition between any two vectors is defined and commutative.

Vectors are depicted in vector theory as oriented line segments with lengths equal to their magnitudes. The area of a triangle formed by vectors will be discussed here. When we try to figure out the area of a triangle, we most often use Heron’s Formula to calculate the value. Vectors can also be used to represent the area of a triangle.

The concept of orthogonality is a generalization of the concept of perpendicularity. When two vectors are perpendicular to each other, they are said to be orthogonal. In other words, the dot product of the two vectors is zero.

Assume that $\overrightarrow{A}$ and $\overrightarrow{B}$ are two linearly independent vectors. We know that the cross-product of two linearly independent vectors yields a non-zero vector that is orthogonal to both.

Let

$\overrightarrow{A}=\overrightarrow{PQ}$

$\overrightarrow{A}=(-2,1,4)-(1,0,1)$

$\overrightarrow{A}=(-3,1,3)$

And

$\overrightarrow{B}=\overrightarrow{PR}$

$\overrightarrow{B}=(7,2,7)-(1,0,1)$

$\overrightarrow{B}=(6,2,6)$

Let $\overrightarrow{C}$ be a non-zero vector orthogonal to the plane through the points $P,Q$ and $R$, then

$\overrightarrow{C}=\overrightarrow{A}\times\overrightarrow{B}$

$=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-3&1&3\\6&2&6\end{vmatrix}$

$=(6-6)\hat{i}-(-18-18)\hat{j}+(-6-6)\hat{k}$

$=0\hat{i}+36\hat{j}-12\hat{k}$

$=<0,36,-12>$

Since it is known that $\overrightarrow{A}$ and $\overrightarrow{B}$ are two sides of a triangle, we also know that the magnitude of the cross-product can be used to calculate the area of the triangle, therefore

Area of the triangle $=\dfrac{1}{2}|\overrightarrow{A}\times \overrightarrow{B}|$

$=\dfrac{1}{2}\sqrt{0^2+36^2+(-12)^2}$

$=\sqrt{1296+144}=\dfrac{1}{2}(12\sqrt{10})$

$=6\sqrt{10}$

## Example

Consider a triangle $ABC$. The values of $\overrightarrow{A},\overrightarrow{B}$ and $\overrightarrow{C}$ are:

$\overrightarrow{A}=5\hat{i}+\hat{j}+3\hat{k}$

$\overrightarrow{B}=7\hat{i}+2\hat{j}+5\hat{k}$

$\overrightarrow{C}=-\hat{i}-3\hat{j}-10\hat{k}$

Find the area of the triangle.

### Solution

Since the area of the triangle is $=\dfrac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$

Now,

$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}$

$=(7\hat{i}+2\hat{j}+5\hat{k})-( 5\hat{i}+\hat{j}+3\hat{k})$

$=2\hat{i}+\hat{j}+2\hat{k}$

And

$\overrightarrow{AC}=\overrightarrow{ C}-\overrightarrow{A}$

$=(-\hat{i}-3\hat{j}-10\hat{k})-( 5\hat{i}+\hat{j}+3\hat{k})$

$=-6\hat{i}-4\hat{j}-13\hat{k}$

Also, $\overrightarrow{AB}\times \overrightarrow{AC}$

$=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&2\\-6&-4&-13\end{vmatrix}$

$=\hat{i}(-13+8)+\hat{j}(-26+12)-(-8+6)\hat{k}$

$=-5\hat{i}-14\hat{j}+2\hat{k}$

$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{(-5)^2+(-14)^2+(2)^2}$

$=\sqrt{25+196+4}$

$=\sqrt{225}=15$

Area of triangle $=\dfrac{15}{2}$.

Images/mathematical drawings are created with GeoGebra.