Clarifying Orthogonal Vectors: Understanding Homework Notation

[FrogPad]

I'm confused on the following questions.

(1) Find a vector that is perpendicular to (v_1,v_2).
(2) Find two vectors that are perpendicular to (v_1,v_2,v_3.

This homework set was written by the professor (it is review) before we actually get into the new material. The notation the book uses is that italics variables names are vectors. However in this homework set the professor uses bold face to denote vectors.

I'm just unsure what the question is asking. Is (v_1,v_2) supposed to be 2 vectors in cartesian coordinates of the third dimension. Or does (v_1,v_2) denote one vector with components v_1 and v_2.

I thought this review was going to be very straightforward, so I waited to the last minute (so I don't have time to ask him).

Anyways, any clarification would be helpful. What do you think it would be? I know that:

\vec v \cdot \vec y = 0
means that the vectors are orthogonal to each other.

and that
\vec v \times \vec y = \vec a
means that \vec a is orthogonal to both \vec v and \vec y.

is that enough knowledge to complete the exercise?

thanks in advance.

 

[benorin]

Are particular values of v_1,v_2, or v_3 given anywhere prior to the exercise? What is the notation used in the section? e.g. are vectors denoted by the use of angle braces (e.g., \vec{v}=\left< 3,4\right>)?

 

[FrogPad]

no values are given to v_1, v_2, or v_3.

The review problems are actually from Strang's Calculus (which I do not own). The professor picked this book and grabbed some review questions from it. The book for our class is Strang's Introdution to Applied Mathematics.

wait... he put that the questions are from Strang's Calculus on the top of the page. Since Strang uses italics as vector notation in the other book, I'm going to just assume that v_1, ... are actually vectors and in R^3.

:) I hope that is right though.

 

[benorin]

I have stewart, what edition, page, and chapter?

 

[HallsofIvy]

I'm confused on the following questions.

(1) Find a vector that is perpendicular to (v_1,v_2).
(2) Find two vectors that are perpendicular to (v_1,v_2,v_3.

This homework set was written by the professor (it is review) before we actually get into the new material. The notation the book uses is that italics variables names are vectors. However in this homework set the professor uses bold face to denote vectors.

I'm just unsure what the question is asking. Is (v_1,v_2) supposed to be 2 vectors in cartesian coordinates of the third dimension.
Or does (v_1,v_2) denote one vector with components v_1 and v_2. [\quote]
It wouldn't make sense to ask for a vector perpendicular to three vectors. I think it is clear that "v_1", "v_2", "v_3" are components of a vector. There are, of course, an infinite number of correct answers to these. Remember that two vectors are perpendicular if and only if their dot product is 0 and pick easy numbers.

I thought this review was going to be very straightforward, so I waited to the last minute (so I don't have time to ask him).

Anyways, any clarification would be helpful. What do you think it would be? I know that:

\vec v \cdot \vec y = 0
means that the vectors are orthogonal to each other.

and that
\vec v \times \vec y = \vec a
means that \vec a is orthogonal to both \vec v and \vec y.

is that enough knowledge to complete the exercise?

thanks in advance.

All you really need is to know that two vectors are perpendicular if and only if their dot product is 0.

For example if I were asked to find 2 vectors both perpendicular to
<1, 3, -4>, I might choose <4, 0, 1> and <3, -1, 0>. Do you see why?

 

[FrogPad]

OK... the homework was not actually due until Wednesday. Anyways I asked the professor today, and the v1,.. are actually components of the vector.

Halls:
I was thinking about using the cross product under the asssumption that v1,v2 were actually vectors. I thought it would be an easy way to compute a vector that is orthogonal to the two. Knowing that they are components makes it a pretty straightforward exercise using the dot product.

Thank you all,