Chapter 11.4, Problem 27E

To determine

To find: The general form of the equation of a place passing through the given three points.

Answer to Problem 27E

$-3x-9y+7z=0$

Explanation of Solution

Given information:

A plane is passing through point $(0,0,0)$ , $(1,2,3)$ ,and $(-2,3,3)$ .

Concepts used:

A plane containing the point $P(x_1,y_1,z_1)$ perpendicular to the normal vector $\text{v}=\langle a,b,c\rangle$

can be represented by standard form of the equation of a plane −

$a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ .

Calculation:

Let plane is passing through the points $P$

$(0,0,0)$ , $Q$

$(1,2,3)$ , and $R$

$(-2,3,3)$ .

So, $x_1=0,y_1=0,z_1=0$ .

To find a normal vector to the plane, take cross product of vector $\overrightarrow{PQ}\text{ and }\overrightarrow{PR}$ .

$\begin{array}{l}\overrightarrow{PQ}=\langle 1-0,2-0,3-0\rangle \\ \overrightarrow{PQ}=\langle 1,2,3\rangle \\ \overrightarrow{PR}=\langle -2-0,3-0,3-0\rangle \\ \overrightarrow{PR}=\langle -2,3,3\rangle \end{array}$

  $\begin{array}{l}\text{n}=\overrightarrow{PQ}\times \overrightarrow{PR}\\ \text{n}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ -2& 3& 3\end{array}\right|\\ \text{n}=i(6-9)-j(3+6)+k(3+4)\\ \text{n}=-3i-9j+7k\\ \text{n}=\langle -3,-9,7\rangle \end{array}$

Line is perpendicular to the normal vector $\text{n}=\langle -3,-9,7\rangle$ .

So, direction numbers are $a=-3,b=-9,c=7$ .

Therefore, general form of the equation of plane will be given by −

$\begin{array}{l}a(x-x_1)+b(y-y_1)+c(z-z_1)=0\\ \Rightarrow -3(x-0)-9(y-0)+7(z-0)=0\\ \Rightarrow -3x-9y+7z=0\end{array}$

Hence, desired equation of plane is $-3x-9y+7z=0$ .