Chapter 8, Problem 42SGA

To determine

To Find: The vector that is perpendicular to the plane containing the points $\left(1,2,3\right)$ , $\left(-4,2,-1\right)$ and $\left(5,-3,0\right)$ .

Answer to Problem 42SGA

The vector is $-20\widehat{i}-40\widehat{j}-20\widehat{i}$ .

Explanation of Solution

Given:

The given points are $\left(1,2,3\right)$ , $\left(-4,2,-1\right)$ and $\left(5,-3,0\right)$ .

Calculation:

Consider the points are,

  $\begin{array}{l}A\left(1,2,3\right)\\ B\left(-4,2,-1\right)\\ C\left(5,-3,0\right)\end{array}$

Consider the vector $\overrightarrow{AB}$ is,

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

Consider the vector $\overrightarrow{BC}$ is,

  $\begin{array}{c}\overrightarrow{BC}=\left(5-\left(-4\right)\right)i+\left(-3-2\right)j+\left(0-\left(-1\right)\right)k\\ =-9i-5j+k\end{array}$

Consider the cross product of the two vectors is,

  $\begin{array}{c}\overrightarrow{AB}\times \overrightarrow{BC}=\left|\begin{array}{ccc}i& j& k\\ -5& 0& -4\\ -9& -5& 1\end{array}\right|\\ =-20\widehat{i}-40\widehat{j}-20\widehat{i}\end{array}$

Thus, the vector is $-20\widehat{i}-40\widehat{j}-20\widehat{i}$ .