Answer to Question #222326 in Analytic Geometry for john

Question

Find the equation of the plane through the points (1,-2,4) ,(3,1,-2) and (2,-1,6)

Answer 1

$\text{The equation of the plane is given by} \\(\overrightarrow{r} - \overrightarrow{a}).\overrightarrow{n}=0 \\\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n} \\\text{where $\overrightarrow{a}=(1,-2,4), \overrightarrow{r} = (x,y,z)$}\\\text{Next, we compute $\overrightarrow{n}$, to this we compute the following vectors and find the vector} \\\text{cross product of the 2 vectors} \\(3,1,-2)-(1,-2,4)=(2,3,-6) \\(2,-1,6)-(3,1,-2) = (-1,-2,8) \\\therefore \overrightarrow{n} = \begin{vmatrix} i&j&k\\ 2&3&-6\\ -1 & -2&8 \end{vmatrix} \\\implies \overrightarrow{n}=12i-10j-k \\\text{Next, we substitute the values of $\overrightarrow{n}$, $\overrightarrow{a}$ and $\overrightarrow{r}$ in equation(1), we have} \\ 12x-10y-z=28$