Chapter 9.6, Problem 31E

To determine

To find: the equation of the plane.

Answer to Problem 31E

The equation of the plane is $12x+4y+3z=12$ .

Explanation of Solution

Given information:

  $a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0.$

Calculation:

Calculate the value of two vector $n=\langle a,b,c\rangle$ by using $n=\overrightarrow{PQ}\times \overrightarrow{PR}$ .

The vector $\overrightarrow{PQ}$ with initial point $P\left(x_{1,}{y}_1{z}_1\right)$ and terminal point $Q\left(x_{2,}{y}_{2,}{z}_2\right)$ is given by $\overrightarrow{PQ}=\left(x_2-x_1\right)i+\left(y_2-y_1\right)j+\left(z_2-z_1\right)k.$

Using the initial point $P\left(1,0,0\right)$ and the terminal point $Q\left(0,3,0\right)$ ,we get

  $\begin{array}{c}\overrightarrow{PQ}=\left(0-1\right)i+\left(3-0\right)j+\left(0-0\right)k.\\ =-i+3j.\end{array}$

Using the initial point $P\left(1,0,0\right)$ and terminal point $R\left(0,0,4\right)$ ,we get

  $\begin{array}{c}\overrightarrow{PR}=\left(0-1\right)i+\left(0-0\right)j+\left(4-0\right)k\\ =-i+4k.\end{array}$

The cross product of the two vectors $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ is given by

  $a\times b=\left(\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right)$ .

Calculate the value of two vectors $a\times b$ with $\overrightarrow{PQ}\times \overrightarrow{PR}$ ,we get

  $\overrightarrow{PQ}\times \overrightarrow{PR}=\left(\begin{array}{ccc}i& j& k\\ -1& 3& 0\\ -1& 0& 4\end{array}\right)$ .

Expending the above determinant along first row, we get

  $\begin{array}{c}\overrightarrow{PQ}\times \overrightarrow{PR}=i\left(12-\left(0\right)\right)-j\left(-4-0\right)+k\left(0-\left(-3\right)\right)\\ =12i+4k+3k.\end{array}$

Thus ,the normal vector to the plane is $n=12i+4j+3k.$

Will calculate,

  $x_0=1$ , $y_0=0,$

  $z_0=0$

  $a=12,$

  $b=4,$

  $c=3.$

On the substitute these values in the equation $a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0.$

  $12\left(x-1\right)+4\left(y-0\right)+3\left(z-0\right)=0.$

On transposing the constant terms to right hand side of the equation ,

  $12x+4y+3z=12.$

Thus, the equation of the plane that crosses the $x$ -axis at $x=1,$ the y-axis at $y=3$ and the $z$ -axis at $z=4$ is $12x+4y+3z=12.$