Question

Find the vector and Cartesian equation of the plane passing through the points with position vectors \[3\overrightarrow i + 4\overrightarrow j + 2\overrightarrow k \], \[2\overrightarrow i - 2\overrightarrow j - \overrightarrow k \] and \[7\overrightarrow i + \overrightarrow k \].

Answer 1

Hint:
Here we need to find the Cartesian and the vector equation of the plane. For that, we will find the cross product of the two vectors to find the direction ratios of the plane. Then we will find the value of the constant term of the term using the given points. we will find the required vector and the Cartesian equation of the plane.

Formula used:
\[\overrightarrow {BC} = \overrightarrow B - \overrightarrow A \]

Complete step by step solution:
Here we need to find the Cartesian and the vector equation of the plane.
Here we have been given three point vectors.
Let \[\overrightarrow A = 3\overrightarrow i + 4\overrightarrow j + 2\overrightarrow k \], \[\overrightarrow B = 2\overrightarrow i - 2\overrightarrow j - \overrightarrow k \] and \[\overrightarrow C = 7\overrightarrow i + \overrightarrow k \].
Now, we will calculate the vector \[\overrightarrow {AC} \] which can be calculated by \[\overrightarrow C - \overrightarrow A \] i.e.
\[\overrightarrow {AC} = \overrightarrow C - \overrightarrow A \]
Now, we will substitute the value of the vectors here.
\[\begin{array}{l} \Rightarrow \overrightarrow {AC} = 7\overrightarrow i + \overrightarrow k - \left( {3\overrightarrow i + 4\overrightarrow j + 2\overrightarrow k } \right)\\ \Rightarrow \overrightarrow {AC} = 7\overrightarrow i + \overrightarrow k - 3\overrightarrow i - 4\overrightarrow j - 2\overrightarrow k \end{array}\]
On subtracting the like terms, we get
\[ \Rightarrow \overrightarrow {AC} = 4\overrightarrow i - 4\overrightarrow j - \overrightarrow k \]
Now, we will calculate the vector \[\overrightarrow {BC} \] which can be calculated by \[\overrightarrow B - \overrightarrow A \] i.e.
\[\overrightarrow {BC} = \overrightarrow B - \overrightarrow A \]
Now, we will substitute the value of the vectors here.
\[\begin{array}{l} \Rightarrow \overrightarrow {BC} = 7\overrightarrow i + \overrightarrow k - \left( {2\overrightarrow i - 2\overrightarrow j - \overrightarrow k } \right)\\ \Rightarrow \overrightarrow {BC} = 7\overrightarrow i + \overrightarrow k - 2\overrightarrow i + 2\overrightarrow j + \overrightarrow k \end{array}\]
On subtracting the like terms, we get
\[ \Rightarrow \overrightarrow {BC} = 5\overrightarrow i - 2\overrightarrow j + 2\overrightarrow k \]
Therefore the directional ratios of normal of required plane will be equal to the cross product of \[\overrightarrow {AC} \] and \[\overrightarrow {BC} \].
Now, we will find the cross product of \[\overrightarrow {AC} \] and \[\overrightarrow {BC} \].
\[ \Rightarrow \overrightarrow {AC} \times \overrightarrow {BC} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\4&{ - 4}&{ - 1}\\5&{ - 2}&2\end{array}} \right|\]
On finding the determinant, we get
\[ \Rightarrow \overrightarrow {AC} \times \overrightarrow {BC} = \widehat i\left( { - 4 \times 2 - \left( { - 2} \right) \times \left( { - 1} \right)} \right) - \widehat j\left( {4 \times 2 - \left( { - 1} \right) \times 5} \right) + \widehat k\left( {4 \times \left( { - 2} \right) - 5 \times \left( { - 4} \right)} \right)\]
On further simplifying the terms, we get
\[\begin{array}{l} \Rightarrow \overrightarrow {AC} \times \overrightarrow {BC} = \widehat i\left( { - 8 - 2} \right) - \widehat j\left( {8 + 5} \right) + \widehat k\left( { - 8 + 20} \right)\\ \Rightarrow \overrightarrow {AC} \times \overrightarrow {BC} = - 10\widehat i - 13\widehat j + 12\widehat k\end{array}\]
Therefore, the direction ratios of the given plane will be -10, -13, 12. We know that the Cartesian equation of a plane is given by
\[ax + by + cz = k\]
Here, \[a\], \[b\] and \[c\] are the direction ratios of the plane.
On substituting the value of the direction ratios, we get
\[ - 10x - 13y + 12z = k\].
Now, we will substitute the coordinates of any one of the given point here. Here, we will substitute the point \[\overrightarrow B = 2\overrightarrow i - 2\overrightarrow j - \overrightarrow k \] or \[\left( {2, - 2, - 1} \right)\]
\[ \Rightarrow - 10 \times 2 - 13 \times - 2 + 12 \times - 1 = k\]
On multiplying the numbers, we get
\[\begin{array}{l} \Rightarrow - 20 + 26 - 12 = k\\ \Rightarrow - 6 = k\end{array}\]

Therefore, the Cartesian equation of a plane is equal to \[ - 10x - 13y + 12z = - 6\]
So the vector equation will equal to \[\overrightarrow r \cdot \left( { - 10\widehat i - 13\widehat j + 12\widehat k} \right) = - 6\].

Note:
Here we have obtained the Cartesian and the vector equation of the plane. Here plane is defined as two dimensional surfaces that extend to infinity. Remember that if we consider any two points lying the plane and if we join them then it will form a straight line.