Question

Show that the points \[A\left( {3,2,1} \right)\], \[B\left( {4,5,5} \right)\], \[C\left( {4,2, - 2} \right)\] and \[D\left( {6,5, - 1} \right)\] are coplanar.

Answer 1

Hint:
Here we will find that the four points given are coplanar by using the definition of a coplanar plane. First, we will write all the points in vector form. We will use the fact that the cross product of the two vectors plane is normal to the plane formed by the two vectors. Then we will prove that the third vector is normal to this cross product. Finally, we will solve the equation to get our desired answer.

Complete step by step solution:
The four points given to us are \[A\left( {3,2,1} \right)\], \[B\left( {4,5,5} \right)\], \[C\left( {4,2, - 2} \right)\] and \[D\left( {6,5, - 1} \right)\].
So the vectors formed by them will be,
\[\vec A = 3\hat i + 2\hat j + \hat k\]
\[\vec B = 4\hat i + 5\hat j + 5\hat k\]
\[\vec C = 4\hat i + 2\hat j - 2\hat k\]
\[\vec D = 6\hat i + 5\hat j - \hat k\]
Now we know that the cross product of the first two vector planes is to be found which is \[A\vec B\] and \[A\vec C\]. We know that \[A\vec B\] is calculated as \[\left( {\vec AB} \right) = \vec B - \vec A\]. Therefore, we get,
\[\left( {A\vec B} \right) = 4\hat i + 5\hat j + 5\hat k - 3\hat i + 2\hat j + \hat k\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \left( {A\vec B} \right) = \hat i + 3\hat j + 4\hat k\]
Next, we will calculate \[A\vec C\].
\[\left( {A\vec C} \right) = \vec C - \vec A\]
Substituting \[\vec C = 4\hat i + 2\hat j - 2\hat k\] and \[\vec A = 3\hat i + 2\hat j + \hat k\] in the above equation, we get
\[\left( {A\vec C} \right) = 4\hat i + 2\hat j - 2\hat k - 3\hat i + 2\hat j + \hat k\]
Adding and subtracting the like terms, we get
\[ \Rightarrow \left( {A\vec C} \right) = \hat i - 3\hat k\]
Now we will find the cross product of two vectors as,
\[A\vec B \times A\vec C = \left( {\hat i + 3\hat j + 4\hat k} \right) \times \left( {\hat i - 3\hat k} \right)\]
\[ \Rightarrow A\vec B \times A\vec C = \hat i \cdot \hat i - \hat i \cdot 3\hat k + 3\hat j \cdot \hat i - 3\hat j \cdot 3\hat k + 4\hat k \cdot i - 4\hat k \cdot 3\hat k\]
Now using the property \[\widehat j \times \widehat j = \widehat i \times \widehat i = \widehat k \times \widehat k = 0\], \[\widehat i \times \widehat k = - \widehat j\], \[\widehat i \times \widehat j = \widehat k\] and \[\widehat k \times \widehat j = \widehat i\], we get
\[ \Rightarrow A\vec B \times A\vec C = 0 + 3\hat j - 3\hat k - 9\hat i + 4\hat j - 12 \cdot 0\]
Adding and subtracting the like terms, we get
\[ \Rightarrow A\vec B \times A\vec C = - 9\hat i + 7\hat j - 3\vec k\]
Now, we will find the vector \[A\vec D\].
\[\left( {A\vec D} \right) = \vec D - \vec A\]
Substituting \[\vec D = 6\hat i + 5\hat j - \hat k\] and \[\vec A = 3\hat i + 2\hat j + \hat k\] in the above equation, we get
\[ \Rightarrow \left( {A\vec D} \right) = \left( {6\hat i + 5\hat j - \hat k} \right) - \left( {3\hat i + 2\hat j + \hat k} \right)\]
Adding and subtracting the terms, we get
\[ \Rightarrow \left( {A\vec D} \right) = 3\hat i + 3\hat j - 2\hat k\]
Now we will find \[A\vec D \cdot \left( {A\vec B \times A\vec C} \right)\] to check for coplanar.
\[A\vec D \cdot \left( {A\vec B \times A\vec C} \right) = \left( {3\hat i + 3\hat j - 2\hat k} \right) \cdot \left( { - 9\hat i + 7\hat j - 3\hat k} \right)\]
\[\begin{array}{l}A\vec D \cdot \left( {A\vec B \times A\vec C} \right) = 3\hat i \cdot \left( { - 9} \right)\hat i + 3\hat i \cdot 7\hat j - 3\hat i \cdot 3\hat k + 3\hat j \cdot \left( { - 9} \right)\hat i + 3\hat j \cdot 7\hat j\\ - 3\hat j \cdot 3\hat k - 2\hat k \cdot \left( { - 9} \right)\hat i - 2\hat k \cdot 7\hat j + 2\hat k \cdot 3\hat k\end{array}\]
Multiplying the terms, we get
\[ \Rightarrow A\vec D \cdot \left( {A\vec B \times A\vec C} \right) = - 27 + 0 + 0 + 21 + 0 + 0 + 6\]
Adding and subtracting the terms, we get
\[ \Rightarrow A\vec D \cdot \left( {A\vec B \times A\vec C} \right) = 0\]

So as we get \[A\vec D \cdot \left( {A\vec B \times A\vec C} \right) = 0\] that means that all four points are coplanar.

Note:
In Cross Product, if the same vectors are multiplied then it yields zero, and when different vectors are multiplied yields the third vector. In Dot product, if the same vectors are multiplied then it yields one, and when different vectors are multiplied it yields zero. Coplanar vectors lie in the same plane in a three-dimensional space. Another method to check for coplanar vectors is if no more than two vectors are linearly independent then all the vectors are coplanar.