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

[VinnyCee]

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?

 

[HallsofIvy]

Yes, that looks good- but where is your answer? Since the problem asks for the set of all vectors perpendicular to [1, 3, -1] your answer should be something like "All vectors satisfying" or "all vectors spanned by". You have already calculated that a vector [x, y, z] in that space must satisfy x+ 3y- z= 0; just say that. You have also calculated form that that z= x+ 3y and got [s, t, s+3t] as a "representative" vector. You could answer "all vectors of the form [s, t, s+ 3t] where s and t can be any real numbers. Finally, since [s, t, s+ 3t]= s[1, 0, 1]+ t[0, 1, 3], you could answer "the subspace spanned by [1, 0, 1] and [0, 1, 3]."

Your picture, of course, would be the plane x+ 3y- z= 0.