Find all vectors in R^3 that are perpendicular to [1; 3; -1]

by VinnyCee
Tags: perpendicular, vectors
 P: 492 THE PROBLEM: The dot product is: $$\overrightarrow{x}\,=\,\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right]$$ $$\overrightarrow{y}\,=\,\left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right]$$ in $$\mathbb{R}^n$$: $$\overrightarrow{x}\,\cdot\,\overrightarrow{y}\,=\,x_1\,y_1\,+\,x_2\,y_2 \,+\,\ldots\,+\,x_n\,y_n$$ If the scalar $\overrightarrow{x}\,\cdot\,\overrightarrow{y}$ is equal to zero, the vectors are perpendicular. Find all vectors in $\mathbb{R}^3$ that are perpendicular to $$\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]$$. Draw a sketch as well. MY WORK SO FAR: $$\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]\,\cdot\,\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\,=\,0$$ $$x\,+\,3\,y\,-\,z\,=\,0$$ $$z\,=\,x\,+\,3\,y$$ Let s = x and t = y $$z\,=\,s\,+\,3\,t$$ $$\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]\,\cdot\,\left[ \begin{array}{c} s \\ t \\ s\,+\,3\,t \end{array} \right]\,=\,0$$ Does the above look right?