Register to reply

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

by VinnyCee
Tags: perpendicular, vectors
Share this thread:
Sep12-06, 01:07 AM
P: 492

The dot product is:

[tex]\overrightarrow{x}\,=\,\left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \end{array} \right][/tex]

[tex]\overrightarrow{y}\,=\,\left[ \begin{array}{c} y_1 \\ y_2 \\ \vdots \\ y_n \end{array} \right][/tex]

in [tex]\mathbb{R}^n[/tex]:

[tex]\overrightarrow{x}\,\cdot\,\overrightarrow{y}\,=\,x_1\,y_1\,+\,x_2\,y_2 \,+\,\ldots\,+\,x_n\,y_n[/tex]

If the scalar [itex]\overrightarrow{x}\,\cdot\,\overrightarrow{y}[/itex] is equal to zero, the vectors are perpendicular.

Find all vectors in [itex]\mathbb{R}^3[/itex] that are perpendicular to
[tex]\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right][/tex].

Draw a sketch as well.


[tex]\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]\,\cdot\,\left[ \begin{array}{c} x \\ y \\ z \end{array} \right]\,=\,0[/tex]



Let s = x and t = y


[tex]\left[ \begin{array}{c} 1 \\ 3 \\ -1 \end{array} \right]\,\cdot\,\left[ \begin{array}{c} s \\ t \\ s\,+\,3\,t \end{array} \right]\,=\,0[/tex]

Does the above look right?
Phys.Org News Partner Science news on
Fungus deadly to AIDS patients found to grow on trees
Canola genome sequence reveals evolutionary 'love triangle'
Scientists uncover clues to role of magnetism in iron-based superconductors
Sep12-06, 09:23 AM
Sci Advisor
PF Gold
P: 39,503
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.

Register to reply

Related Discussions
Find two perpendicular vectors which are perpendicular to another Calculus 14
Equation of a line perpendicular to 2 vectors Calculus & Beyond Homework 3
Find slope of 2 perpendicular lines Precalculus Mathematics Homework 11
Find plane, given a line & perpendicular plane Calculus & Beyond Homework 2
Perpendicular 3Dimensional Vectors Introductory Physics Homework 3