Q 25. - Calculus

Question

Find the equations of the planes determined by the given conditions. 

The plane contains the points $(2, 4, 3), (3,-5, 0)$ and $(-4, 1, 6)$

Step-by-Step Solution

Answer

The equation of the plane that contains the points $P=(2,4,3), Q=(3,-5,0)$ and $R=(-4,1,6)$ is $12 x-5 y+19 z-61=0$

Step 1: Given information

The plane that contains the points $P=(2,4,3), Q=(3,-5,0)$ and $R=(-4,1,6)$

Step 2: Calculation

The goal is to determine the plane equation that is determined by the given conditions.

To find the plane's equation, first, demonstrate that the points are non-collinear.

The vector $\overrightarrow{PQ}$ is:

$$\begin{gathered} \overrightarrow{PQ}=⟨3-2,-5-4,0-3⟩ \\ =⟨1,-9,-3⟩ \end{gathered}$$

The vector $\overrightarrow{QR}$ is:

$$\begin{gathered} \overrightarrow{OR}=(-4-3,1-(-5),6-0) \\ =⟨-7,6,6⟩ \end{gathered}$$

The vector $\overline{RP}$ is:

$$\begin{gathered} \overline{RP}=⟨2-(-4),4-1,3-6⟩ \\ =⟨6,3,-3⟩ \end{gathered}$$

Because neither of the vectors is a scalar multiple of the other, they are non-collinear.

Step 3: Calculation

Calculate the normal vector by using the cross-product of any two vectors to determine the plane's equation.

The normal vector $\mathbf{N}=\overrightarrow{PQ}\times \overrightarrow{RP}$ is given by:

$$\begin{gathered} \mathbf{N}=\left|\begin{array}{c}ijk\\ 1-9-3\\ 63-3\end{array}\right| \\ =36\mathbf{i}-15\mathbf{j}+57\mathbf{k} \\ =⟨36,-15,57⟩ \end{gathered}$$

The equation of the plane is:

$$\begin{gathered} 36(x-2)-15(y-4)+57(z-3)=0 \\ 36x-15y+57z-72+60-171=0\text{ (Simplify) } \\ 36x-15y+57z-183=0\text{ (Combine like terms) } \\ 12x-5y+19z-61=0\text{ (Divide by }3\text{ ) } \end{gathered}$$

The equation of the plane that contains the points $P=(2,4,3), Q=(3,-5,0)$ and $R=(-4,1,6)$ is $12 x-5 y+19 z-61=0$