Question

How do you find all unit vectors that are orthogonal to the plane through the points P = (3,-3,0), Q = (5,-1,2), and R = (5,-1,6)?

Answer 1

Hint: This question is from the topic of algebra. In solving this question, we will first find the vector using the points P and Q. After that, we will find the vector using the points Q and R. From using the vectors PQ and QR, we will find out a vector which is orthogonal to the plane. After that, we will find out the magnitude of the resultant vector. After that, we will divide the resultant vector with the magnitude, and then we will get the unit vector.

Complete step by step solution:
Let us solve this question.
In this question, we have asked to find the unit the vector that is orthogonal to the plane through the points P = (3,-3,0), Q = (5,-1,2), and R = (5,-1,6)
The vector PQ can be written as
\[\overrightarrow{PQ}=\left( 5-3 \right)\overset{\wedge }{\mathop{i}}\,+\left( -1-\left( -3 \right) \right)\overset{\wedge }{\mathop{j}}\,+\left( 2-0 \right)\overset{\wedge }{\mathop{k}}\,=2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\,\]
The vector QR can be written as
\[\overrightarrow{QR}=\left( 5-5 \right)\overset{\wedge }{\mathop{i}}\,+\left( -1-\left( -1 \right) \right)\overset{\wedge }{\mathop{j}}\,+\left( 6-2 \right)\overset{\wedge }{\mathop{k}}\,=0\overset{\wedge }{\mathop{i}}\,+0\overset{\wedge }{\mathop{j}}\,+4\overset{\wedge }{\mathop{k}}\,\]
As we know that the points P, Q, and R are in the same plane, then we can say that the vectors \[\overrightarrow{PQ}\] and \[\overrightarrow{QR}\] will also be in the same plane.
Now, we can say that the vector product of these vectors will be orthogonal or perpendicular to these vectors that are \[\overrightarrow{PQ}\] and \[\overrightarrow{QR}\].
So, let the vector product be \[\overrightarrow{n}\]. Hence, the vector product will be
\[\overrightarrow{n}=\overrightarrow{PQ}\times \overrightarrow{QR}\]
Or, we can write
\[\overrightarrow{n}=\left( 2\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,+2\overset{\wedge }{\mathop{k}}\, \right)\times \left( 0\overset{\wedge }{\mathop{i}}\,+0\overset{\wedge }{\mathop{j}}\,+4\overset{\wedge }{\mathop{k}}\, \right)\]
This can be written as
\[\overrightarrow{n}=\left| \begin{matrix}
   \overset{\wedge }{\mathop{i}}\, & \overset{\wedge }{\mathop{j}}\, & \overset{\wedge }{\mathop{k}}\, \\
   2 & 2 & 2 \\
   0 & 0 & 4 \\
\end{matrix} \right|\]
The above determinant can also be written as
\[\overrightarrow{n}=\left( 2\times 4-2\times 0 \right)\overset{\wedge }{\mathop{i}}\,-\left( 2\times 4-2\times 0 \right)\overset{\wedge }{\mathop{j}}\,+\left( 2\times 0-2\times 0 \right)\overset{\wedge }{\mathop{k}}\,=8\overset{\wedge }{\mathop{i}}\,-8\overset{\wedge }{\mathop{j}}\,\]
The above equation can also be written as
\[\overrightarrow{n}=8\left( \overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\, \right)\]
The magnitude of the vector \[\overrightarrow{n}\] will be
\[\left| \overrightarrow{n} \right|=8\sqrt{{{\left( \overset{\wedge }{\mathop{i}}\, \right)}^{2}}+{{\left( \overset{\wedge }{\mathop{j}}\, \right)}^{2}}}=8\sqrt{2}\]
Hence, the unit vector that is orthogonal to the plane having points P, Q, and R will be \[\overset{\wedge }{\mathop{n}}\,=\dfrac{\overrightarrow{n}}{\left| \overrightarrow{n} \right|}=\dfrac{8\left( \overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\, \right)}{8\sqrt{2}}=\dfrac{1}{\sqrt{2}}\left( \overset{\wedge }{\mathop{i}}\,-\overset{\wedge }{\mathop{j}}\, \right)\]

Note:
We should have a better knowledge in the topic of vectors to solve this type of question easily. We should know how to find the vector from the two given points. Let us understand from the following:
Let the points be \[A=\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)\] and \[B=\left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)\], then the vector AB will be
\[\overrightarrow{AB}=\left( {{x}_{2}}-{{x}_{1}} \right)\overset{\wedge }{\mathop{i}}\,+\left( {{y}_{2}}-{{y}_{1}} \right)\overset{\wedge }{\mathop{j}}\,+\left( {{z}_{2}}-{{z}_{1}} \right)\overset{\wedge }{\mathop{k}}\,\]
We should know how to find the vector product from the two given vectors. Let us understand this from the following:
Suppose, we have two vectors such as \[\left( {{x}_{1}}\overset{\wedge }{\mathop{i}}\,+{{y}_{1}}\overset{\wedge }{\mathop{j}}\,+{{z}_{1}}\overset{\wedge }{\mathop{k}}\, \right)\] and \[\left( {{x}_{2}}\overset{\wedge }{\mathop{i}}\,+{{y}_{2}}\overset{\wedge }{\mathop{j}}\,+{{z}_{2}}\overset{\wedge }{\mathop{k}}\, \right)\]
Then, the vector product will be \[\left( {{x}_{1}}\overset{\wedge }{\mathop{i}}\,+{{y}_{1}}\overset{\wedge }{\mathop{j}}\,+{{z}_{1}}\overset{\wedge }{\mathop{k}}\, \right)\times \left( {{x}_{2}}\overset{\wedge }{\mathop{i}}\,+{{y}_{2}}\overset{\wedge }{\mathop{j}}\,+{{z}_{2}}\overset{\wedge }{\mathop{k}}\, \right)\] which can also be written as
\[\left| \begin{matrix}
   \overset{\wedge }{\mathop{i}}\, & \overset{\wedge }{\mathop{j}}\, & \overset{\wedge }{\mathop{k}}\, \\
   {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\
   {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\
\end{matrix} \right|=\left( {{y}_{1}}{{z}_{2}}-{{z}_{1}}{{y}_{2}} \right)\overset{\wedge }{\mathop{i}}\,-\left( {{x}_{1}}{{z}_{2}}-{{x}_{2}}{{z}_{1}} \right)\overset{\wedge }{\mathop{j}}\,+\left( {{x}_{1}}{{y}_{2}}-{{x}_{2}}{{y}_{1}} \right)\overset{\wedge }{\mathop{k}}\,\]