Question

Find the vector and Cartesian equations of the plane passing through the points (2, 2, -1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of the plane passing through (4, 3, 1) and parallel to the above plane obtained.

Answer 1

Consider the given three points (2, 2, -1), (3, 4, 2) and (7, 0, 6) as $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ respectively. Now, to find the Cartesian equation of the plane passing through these three points use the determinant formula $\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\\ \end{matrix} \right|=0$ and expand it to get the answer. Now, assume the obtained Cartesian equation of the plane as $ax+by+cz=d$, where a, b, c and d are constants. To find the vector equation of the plane use the relation $\overrightarrow{r}.\left( a\hat{i}+b\hat{j}+z\hat{k} \right)=d$ where $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$. Finally, to find the parallel plane assume it as $ax+by+cz=k$ in the Cartesian form and substitute the point (4, 3, 1) to find the value of k. Once the value of k is found use the relation $\overrightarrow{r}.\left( a\hat{i}+b\hat{j}+z\hat{k} \right)=k$ to get the vector equation.

Complete step-by-step answer:
Here we have been provided with three points (2, 2, -1), (3, 4, 2) and (7, 0, 6) and we are asked to determine the vector and Cartesian equation of the plane passing through these three points. Also we need to find the equation of another plane that will parallel to the above plane and will pass through the point (4, 3, 1).
(i) Now, let us assume the three points (2, 2, -1), (3, 4, 2) and (7, 0, 6) as $\left( {{x}_{1}},{{y}_{1}} \right)$, $\left( {{x}_{2}},{{y}_{2}} \right)$ and $\left( {{x}_{3}},{{y}_{3}} \right)$ respectively. So the Cartesian equation of the plane passing through the given three points is given by the determinant formula $\left| \begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\\ \end{matrix} \right|=0$. Therefore, substituting the given values in the determinant we get,

& \Rightarrow \left| \begin{matrix} x-2 & y-2 & z+1 \\\ 3-2 & 4-2 & 2+1 \\\ 7-2 & 0-2 & 6+1 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow \left| \begin{matrix} x-2 & y-2 & z+1 \\\ 1 & 2 & 3 \\\ 5 & -2 & 7 \\\ \end{matrix} \right|=0 \\\ \end{aligned}$$ On expanding along the first row we get, $$\begin{aligned} & \Rightarrow \left( x-2 \right)\left( 14+6 \right)-\left( y-2 \right)\left( 7-15 \right)+\left( z+1 \right)\left( -2-10 \right)=0 \\\ & \Rightarrow 20\left( x-2 \right)+8\left( y-2 \right)-12\left( z+1 \right)=0 \\\ & \therefore 5x+2y-3z=17 \\\ \end{aligned}$$ Therefore, the above equation is the required Cartesian equation of the plane passing through the given three points. (ii) Now, the vector equation of the Cartesian form of a plane $ax+by+cz=d$ is given by the relation $\overrightarrow{r}.\left( a\hat{i}+b\hat{j}+z\hat{k} \right)=d$ where $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$. Here $\overrightarrow{r}$ is the position vector of the point (x, y, z) lying on the plane. Therefore the required vector equation of the plane is given as: - $$\begin{aligned} & \Rightarrow \overrightarrow{r}.\left( 5\hat{i}+2\hat{j}-3\hat{k} \right)=17 \\\ & \therefore \left( x\hat{i}+y\hat{j}+z\hat{k} \right).\left( 5\hat{i}+2\hat{j}-3\hat{k} \right)=17 \\\ \end{aligned}$$ Therefore, the above equation is the required vector equation of the plane passing through the given three points. (iii) Finally, we have to find the vector equation of the plane that passes through the point (4, 3, 1) and is parallel to the plane $$5x+2y-3z=17$$. Since the required plane is parallel to the plane obtained above so their coefficients of x, y and z will be equal and only the constant term in the R.H.S will be different. Therefore, the required equation of the plane can be assumed as $5x+2y-3z=k$. This plane will satisfy the point (4, 3, 1) because it is passing through it, so substituting the values we get, $\begin{aligned} & \Rightarrow 5\left( 4 \right)+2\left( 3 \right)-3\left( 1 \right)=k \\\ & \Rightarrow k=20+6-3 \\\ & \Rightarrow k=23 \\\ \end{aligned}$ So, the Cartesian equation of the plane will be given as $5x+2y-3z=23$. Therefore, the required vector equation of the plane will be given by the relation: - $$\begin{aligned} & \Rightarrow \overrightarrow{r}.\left( 5\hat{i}+2\hat{j}-3\hat{k} \right)=23 \\\ & \therefore \left( x\hat{i}+y\hat{j}+z\hat{k} \right).\left( 5\hat{i}+2\hat{j}-3\hat{k} \right)=23 \\\ \end{aligned}$$ Therefore, the above equation is the vector form of the plane which passes through the point (4, 3, 1) and is parallel to the plane $$5x+2y-3z=17$$. **Note:** Note that you can remember the formula of the vector equation of a plane passing through the points having position vectors as $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$ given by the relation $\left( \overrightarrow{r}-\overrightarrow{a} \right).\left[ \left( \overrightarrow{b}-\overrightarrow{a} \right)\times \left( \overrightarrow{c}-\overrightarrow{a} \right) \right]=0$. Here the dot (.) symbol is the dot product and the cross $\left( \times \right)$ symbol is the cross product. The above process we have applied is nothing but the simplified form of the mentioned formula.