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

In summary, the dot product of two vectors in \mathbb{R}^n can be found by multiplying the corresponding components and adding them together. If the dot product is equal to zero, the vectors are perpendicular. For a vector in \mathbb{R}^3, [1, 3, -1], the set of all vectors perpendicular to it can be described as all vectors of the form [s, t, s+3t] where s and t are any real numbers, or the subspace spanned by [1, 0, 1] and [0, 1, 3]. This can be visualized as the plane x+3y-z=0.
  • #1
VinnyCee
489
0
THE PROBLEM:

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.


MY WORK SO FAR:

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

[tex]x\,+\,3\,y\,-\,z\,=\,0[/tex]

[tex]z\,=\,x\,+\,3\,y[/tex]

Let s = x and t = y

[tex]z\,=\,s\,+\,3\,t[/tex]

[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?
 
Last edited:
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  • #2
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.
 
  • #3


Yes, your work looks correct so far. To find all vectors perpendicular to [1; 3; -1], we can choose any values for s and t and plug them into the equation z = s + 3t to get different vectors. For example, if we let s = 0 and t = 1, we get the vector [0; 1; 3]. If we let s = 1 and t = 0, we get the vector [1; 0; 1]. These are just two examples, but there are infinitely many vectors that are perpendicular to [1; 3; -1].

As for the sketch, we can visualize the vector [1; 3; -1] as a line in 3-dimensional space. Any vector perpendicular to this line will be a line that intersects it at a right angle. This means that the perpendicular vectors will lie in a plane that is perpendicular to the line [1; 3; -1]. So, we can draw a plane in 3-dimensional space and any vector that lies on this plane will be perpendicular to [1; 3; -1]. This plane can be visualized as a flat surface that is perpendicular to the line [1; 3; -1].
 

1. What is R^3?

R^3 refers to the three-dimensional real coordinate space, where each point is represented by three numbers (x, y, z). It is also known as the Cartesian coordinate system.

2. How do you determine if two vectors are perpendicular?

Two vectors are perpendicular if their dot product is equal to 0. In other words, the angle between the two vectors is 90 degrees.

3. What is the process for finding perpendicular vectors in R^3?

To find perpendicular vectors in R^3, we can use the cross product. We take the given vector [1; 3; -1] and find the cross product with any other vector in R^3. The resulting vector will be perpendicular to the given vector.

4. Can there be more than one vector that is perpendicular to [1; 3; -1]?

Yes, there can be an infinite number of vectors that are perpendicular to [1; 3; -1]. This is because any vector that lies in the plane perpendicular to [1;3;-1] will also be perpendicular to it.

5. How can perpendicular vectors be useful in science?

Perpendicular vectors are useful in many scientific applications, such as physics and engineering. They can be used to represent forces, velocities, and other physical quantities that act at right angles to each other. They are also important in vector calculus and in solving problems involving three-dimensional geometry.

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