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?
 PF Patron Sci Advisor Thanks Emeritus P: 38,395 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.

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