Chapter 8.4, Problem 31E

To determine

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

  $\left(0,0,1\right),\left(1,0,1\right),\left(-1,-1,-1\right)$ .

Answer to Problem 31E

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

Explanation of Solution

Given information:

The points containing the points $\left(0,0,1\right),\left(1,0,1\right),\left(-1,-1,-1\right)$ .

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

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

Now,

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

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,

  $\begin{array}{c}\overrightarrow{d}=A\overrightarrow{B}\times A\overrightarrow{C}\\ =\left|\begin{array}{c}ijk\\ 100\\ -1-1-2\end{array}\right|\\ =i\left(-2\times 0-\left(-1\times 0\right)\right)-j\left(-2\times 1-\left(-1\times 0\right)\right)+k\left(-1\times 1-\left(-1\times 0\right)\right)\\ =2j-k\end{array}$

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