91影视

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

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 $\stackrel{鈫�}{PQ}$is:

$$\begin{gathered} \stackrel{鈫�}{PQ}=鉄�3-2,-5-4,0-3鉄� \\ =鉄�1,-9,-3鉄� \end{gathered}$$

The vector $\stackrel{鈫�}{QR}$is:

$$\begin{gathered} \stackrel{鈫�}{OR}=(-4-3,1-(-5),6-0) \\ =鉄�-7,6,6鉄� \end{gathered}$$

The vector $\stackrel{炉}{RP}$is:

$$\begin{gathered} \stackrel{炉}{RP}=鉄�2-(-4),4-1,3-6鉄� \\ =鉄�6,3,-3鉄� \end{gathered}$$

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

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

The normal vector $\mathbf{N}=\stackrel{鈫�}{PQ}脳\stackrel{鈫�}{RP}$is given by:

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

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 $12x-5y+19z-61=0$