Question

question below :)

Answer 1

@ybarrap

Answer 2

So, we need two \(new \) vectors. Each of these two new vectors must be perpendicular to r and s. Is that how you understand the question?

Answer 3

yeah i think so..
i used the dot product to calculate for that..
but i end with two equations then i tried to solve them using elimination but when i did that, i eliminated all of the variables ...

Answer 4

One vector is very easy and is perpendicular to every vector, the zero vector. Because,

$$
\vec{t}\bullet<0,0,0>=<0,0,0>
$$
For any \(\vec{t}\).

So we just need one more vector. See that?

Answer 5

yeah.. okay...
but how do we find the other one?

Answer 6

Do you know how to compute the cross product?

Answer 7

yes, i know that, i learn that during summer
but my teacher did not teach our class yet about cross product :/

Answer 8

even thoguh i used the cross product i still get 0,0,0

Answer 9

That is your 3rd vector. The result of the cross product of r and s is a vector. This vector is perpendicular to both:

$$
\large
\vec r \times \vec s= <1,2,-1>\times<-2,-4,2>\\
=\begin{vmatrix}
\hat i&\hat j &\hat k\\
1&2 &-1\\
-2&-4 &2
\\\end{vmatrix}\\
=\hat i[(2)\times (2)-(-4)\times (-1)]\\
\quad-\hat j[(1)\times (2)-(-2)\times (-1)]\\
\qquad +\hat k[(1)\times (-4)-(-2)\times (2)]\\
=<0,0,0>
$$
Unfortunately, this also gives us the zero vector.

Answer 10

yeah :/
anyways..i just looked the answer at the book and it says.. "infinitely many vectors are perpendicular to ... the given vectors"

Answer 11

I suppose that makes sense because if you extend to other dimensions, their are an infinite number of zero vectors. For example,
$$
\vec r=<1,2,-1,0>\bullet<0,0,0,0>=\bf {\vec 0}\\
=<1,2,-1,0,0>\bullet<0,0,0,0,0>=\bf {\vec 0}
$$etc

Answer 12

*there

Answer 13

One thing that we should have been obvious from the beginning is that
$$
\vec s=-2\vec r
$$
That is, r and s are the same vectors! One is just a scalar multiple of the other. So given any single vector, take a vector perpendicular to it. Then each rotation of this new vector will also be perpendicular to it. Since there are an infinite number of rotations possible, there are an infinite number of perpendicular vectors.

Finally we come to the right conclusion!