Chapter 11, Problem 64RE

To determine

Find the general form of the equation of the plane with the given characteristics.

Answer to Problem 64RE

The general equation of the plane is $–2y+5z=14$ .

Explanation of Solution

Given:

The points

  $\left(–1,3,4\right),\left(4,–2,2\right),\left(2,8,6\right)$

Consider the points $\left(x_1,y_1,z_1\right),\left(x_2,y_2,z_2\right),\left(x_3,y_3,z_3\right)$ then the general equation of the plane passing through these points is:

  $a\left(x–x_1\right)+b\left(y–y_1\right)+c\left(z–z_1\right)=0$

To find the equation of the plane, a point in the plane and a vector that is normal to the plane. To obtain a normal vector, use the cross product of vectors u and v extending from the point $\left(x_1,y_1,z_1\right)$ to the points $\left(x_2,y_2,z_2\right),\left(x_3,y_3,z_3\right)$ .The component forms of u and v are

  ${\begin{array}{l}\text{u}=\langle x_2–x_1,y_2–y_1,z_2–z_1\rangle \hfill \\ \text{v}=\langle x_3–x_1,y_3–y_1,z_3–z_1\rangle \hfill \end{array}}_{}$

Now find the cross product of u and v that is the vector n.

Then by substituting the values of the vector n and the initial point $\left(x_1,y_1,z_1\right)$ in the equation of plane, and get the general equation of plane.

Consider the following points

  $\left(–1,3,4\right),\left(4,–2,2\right),\left(2,8,6\right)$

To obtain a normal vector, use the cross product of vectors u and v extending from the point $\left(–1,3,4\right)$ to the points $\left(4,–2,2\right),\left(2,8,6\right)$ .The component forms of u and v are

  $\begin{array}{l}\text{u}=\langle 4+1,–2–3,2–4\rangle \hfill \\ =\langle 5,–5,–2\rangle \hfill \\ \text{v}=\langle 2+1,8–3,6–4\rangle \hfill \\ =\langle 3,5,2\rangle \hfill \end{array}$

Now find the cross product of u and v that is the vector n.

  $\begin{array}{l}n=u\times v\\ =\left|\begin{array}{ccc}i& j& k\\ 5& -5& -2\\ 3& 5& 2\end{array}\right|\\ =i[(-5)(2)-5(-2)]-j[5(2)-3(-2)]+k[5(5)-3(-5)]\end{array}$

Simplify,

  $\begin{array}{l}=i[-10+10]-j[10+6]+k[25+15]\hfill \\ =-16j+40k\hfill \\ =\langle 0,-16,40\rangle \hfill \end{array}$

Therefore,

  $\langle 0,–16,40\rangle =\langle a,b,c\rangle$

The vector n is normal to the given plane . Use the direction numbers for n and the initial point

  $\left(x_1,y_1,z_1\right)=\left(–1,3,4\right)$ . Determine an equation of the plane to be

  $\begin{array}{l}\begin{array}{l}a\left(x–x_1\right)+b\left(y–y_1\right)+c\left(z–z_1\right)=0\hfill \\ 0\left(x+1\right)–16\left(y–3\right)+40\left(z–4\right)=0\hfill \\ –16y+48+40z–160=0\hfill \end{array} \\ –2y+5z=14\end{array}$

Hence the general equation of the plane is $–2y+5z=14$ .