Question

How do you determine scalar, vector, and parametric equations for the plane that passes through the points $A\left( {1, - 2,0} \right)$, $B\left( {1, - 2,2} \right)$ and $C\left( {0,3,2} \right)$.

Answer 1

Given the points through which the plane passes. We will find the scalar equation of the plane by first finding the difference between the coordinates of the points. Then substitute the resultant coordinates into the general equation of the plane to determine the scalar equation. Then calculate the cross product of two vectors using the determinant and find the normal vector perpendicular to the cross product. Then the parametric equation is calculated by substituting $t$ for $x$.
Formula used:
The formula for finding the vector form of the equation of plane is given as:
$r \cdot n = a \cdot n$
Where $r = xi + yj + zk$, $n$ is the normal vector, and $a$ is any vector point on the plane.
The cross product of two vectors $A = {a_1}i + {a_2}j + {a_3}k$ and $B = {b_1}i + {b_2}j + {b_3}k$ is given by:
${c_1} = {a_2}{b_3} - {a_3}{b_2}$
${c_2} = {a_3}{b_1} - {a_1}{b_3}$
${c_3} = {a_1}{b_2} - {a_2}{b_1}$

Complete step-by-step solution:
First, we will find the vector $\overrightarrow {AB}$ by subtracting the vector point $A$ from vector point $B$
$\Rightarrow \overrightarrow {AB} = \left( {i - 2j + 2k} \right) - \left( {i - 2j + 0k} \right)$
$\Rightarrow \overrightarrow {AB} = 0i + 0j + 2k$
Now, we will find the vector $\overrightarrow {AC}$ by subtracting the vector point $A$ from vector point $C$
$\Rightarrow \overrightarrow {AC} = \left( {0i + 3j + 2k} \right) - \left( {i - 2j + 0k} \right)$
On simplifying the equation, we get:
$\Rightarrow \overrightarrow {AC} = - i + 5j + 2k$
Now, we will find the normal vector which is perpendicular to the vectors $\overrightarrow {AB}$ and $\overrightarrow {AC}$ by computing the cross product of these two vectors, $n = \overrightarrow {AB} \times \overrightarrow {AC}$.
$n = \left( {0i + 0j + 2k} \right) \times \left( { - i + 5j + 2k} \right)$
Apply the formula of cross product of two vectors, we get:
$\Rightarrow {c_1} = 0 \times 5 - 2 \times 5$
$\Rightarrow {c_1} = 0 - 10$
$\Rightarrow {c_1} = - 10$
Now, we will find the coordinates for ${c_2}$:
$\Rightarrow {c_2} = 2 \times \left( { - 1} \right) - 0 \times 2$
$\Rightarrow {c_2} = - 2 - 0$
$\Rightarrow {c_2} = - 2$
Now compute the coordinate for ${c_3}$:
$\Rightarrow {c_3} = 0 \times 5 - 0 \times \left( { - 1} \right)$
$\Rightarrow {c_3} = 0 - 0$
$\Rightarrow {c_3} = 0$
Therefore, the coordinates for a normal vector $n$ is written as $\left( { - 10, - 2,0} \right)$. The vector equation of the plane is $- 10i - 2j$
Then, we will substitute the values into the equation $r \cdot n = a \cdot n$
\Rightarrow \left( {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 10} \\\ { - 2} \\\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\\ 3 \\\ 2 \end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}} { - 10} \\\ { - 2} \\\ 0 \end{array}} \right)
Compute the dot product of
$\Rightarrow - 10\left( x \right) + y\left( { - 2} \right) + z\left( 0 \right) = - 10\left( 0 \right) + 3\left( { - 2} \right) + 2\left( 0 \right)$
On simplifying the equation, we get:
$\Rightarrow - 10x - 2y + 0 = - 6$
$\Rightarrow - 2\left( {5x + y} \right) = - 6$
On dividing both sides by $- 2$, we get:
$\Rightarrow \dfrac{{ - 2\left( {5x + y} \right)}}{{ - 2}} = \dfrac{{ - 6}}{{ - 2}}$
$\Rightarrow 5x + y = 3$
Therefore, the scalar equation of the plane is $5x + y = 3$.
Now, we will get the parametric equation by substituting $x = t$ into the scalar equation.
$\Rightarrow 5t + y = 3$
Solve the equation for $y$.
$\Rightarrow y = 3 - 5t$
Therefore, the parametric equations of the plane are:
\left\\{ x = t \\\ y = 3 - 5t \\\ z = 0 \\\ \right.
Hence the vector equation of the plane is $- 10i - 2j$, the scalar equation of the plane is $5x + y = 3$ and the parametric equations are
$x = t$
$y = 3 - 5t$
$z = 0$

Note: In such types of questions students mainly get confused in applying the formula. As they don't know which formula they have to apply. So when the coordinates through which the pane is passing, then the equations can be calculated by finding the cross product of two points. The students make mistakes while substituting the coordinates into the vector equation of the plane.