Chapter 8.4, Problem 29E

To determine

To find:The vector perpendicular to the plane containing the point:

  $\left(0,-2,2\right),\left(1,2,-3\right),\left(4,0,-1\right)$ .

Answer to Problem 29E

The vector that is perpendicular to the plane is $-2i-17j-14k$ .

Explanation of Solution

Given information:

The plane containing the point $\left(0,-2,2\right),\left(1,2,-3\right),\left(4,0,-1\right)$ .

Let the point of $A,B,C$ is $\left(0,-2,2\right),\left(1,2,-3\right),\left(4,0,-1\right)$ . The position vector is given by,

  $\begin{array}{c}A\overrightarrow{B}=\left(1-0\right)i+\left(2+2\right)j+\left(-3-2\right)k\\ =i+4j-5k\end{array}$

Now,

  $\begin{array}{c}A\overrightarrow{C}=\left(4-0\right)i+\left(0+2\right)j+\left(-1-2\right)k\\ =4i+2j-3k\end{array}$

Let the vector in the plane is $\overrightarrow{C}$ . Then the coordinates of this vectors are,

  $xA\overrightarrow{B}+yA\overrightarrow{C}$ .

Now if any vector say $\overrightarrow{d}$ is perpendicular to $\overrightarrow{C}$ then,

  $\overrightarrow{d}\perp A\overrightarrow{B}$ and also $\overrightarrow{d}\perp A\overrightarrow{C}$ .

Now the vector obtained by cross product of position vector is perpendicular to both the vectors.

Hence,

  $\begin{array}{c}\overrightarrow{d}=A\overrightarrow{B}\times A\overrightarrow{C}\\ =\left|\begin{array}{c}ijk\\ 14-5\\ 42-3\end{array}\right|\\ =i\left(4\times -3-\left(-5\times 2\right)\right)-j\left(-3\times 1-\left(-5\times 4\right)\right)+k\left(2\times 1-\left(4\times 4\right)\right)\\ =-2i-17j-14k\end{array}$

Therefore, the vector that is perpendicular to the plane is $-2i-17j-14k$ .