vectors problems

**quah13579** · September 2nd 2009, 07:19 PM

Given the vectors u = (1, 2, 3, 4), v = (−1, 2,−2, 1),w = (0, 2,−2, 2), compute the following:

(a) the length (norm) of each of these vectors (give exact answer);
(b) the angle (in radians) between all pairs of these vectors (3 significant figures);
(c) the unit vector pointing in the opposite direction for each of these vectors.

my revision,any idea how to do those thanks.

my answer for (a) is
u = [1 2 3 4];
>> v = [-1 2 -2 1];
>> w = [0 2 -2 2];
>> norm(u)
ans =
5.4772
>> norm(v)
ans =
3.1623
>> norm(w)
ans =
3.4641

**Bruno J.** · September 2nd 2009, 08:41 PM

Your answers to (a) seem correct, but you are asked an exact answer. So for the first one you should give $\sqrt 30$ , for example.

For (b), use the fact that $u \bullet v = ||u||\ ||v|| \cos \theta$ where $\theta$ is the angle between the two and $\: \bullet \:$ is the dot product.

For (c), can you find a vector of length one pointing in the same direction as each of your vectors? If you can do that then you should have no trouble finding the one which points in the opposite direction.

**chug1** · September 2nd 2009, 09:09 PM

Well the length of a vector [a,b,c,...] is
$ \sqrt[]{{a^2}+{b^2}+{c^2}+...} $
so the legnth of u would be
$\sqrt[]{{1^2}+{2^2}+{3^2}+{4^2}}$
$=\sqrt[]{{1}+{4}+{9}+{16}}$
$=\sqrt[]{30}$
which is what you got, you just need to provide it as a squareroot for the "exact answer"

Sorry i can't help with (b), but for (c) a vector in the opposite direction has the negative of the components of the original vector. e.g. the opposite direction to [a,b,c,d] is [-a,-b,-c,-d], so the opposite direction to w=[0,2,-2,2] is [0,-2,2,-2]. A unit vector in this direction has a length of 1, so it is some positive multiple k of [0,-2,2,-2] which has a length of 1. Calculate the length using the above formula.
$1=k\:\sqrt[]{{0^2}+{(-2)^2}+{2^2}+{(-2)^2}}$
$1=k\:\sqrt[]{0+4+4+4}$
$1=k\:\sqrt[]{12}$
$k=\frac{1}{\sqrt[]{12}}$
so the unit vector in the direction opposite to w=[0,2,-2,2] is
$ \frac{1}{\sqrt[]{12}}\:\left[0,-2,2,-2 \right] $

Hope this helps

**quah13579** · September 3rd 2009, 12:41 AM

Originally Posted by chug1

Well the length of a vector [a,b,c,...] is
$ \sqrt[]{{a^2}+{b^2}+{c^2}+...} $
so the legnth of u would be
$\sqrt[]{{1^2}+{2^2}+{3^2}+{4^2}}$
$=\sqrt[]{{1}+{4}+{9}+{16}}$
$=\sqrt[]{30}$
which is what you got, you just need to provide it as a squareroot for the "exact answer"

Sorry i can't help with (b), but for (c) a vector in the opposite direction has the negative of the components of the original vector. e.g. the opposite direction to [a,b,c,d] is [-a,-b,-c,-d], so the opposite direction to w=[0,2,-2,2] is [0,-2,2,-2]. A unit vector in this direction has a length of 1, so it is some positive multiple k of [0,-2,2,-2] which has a length of 1. Calculate the length using the above formula.
$1=k\:\sqrt[]{{0^2}+{(-2)^2}+{2^2}+{(-2)^2}}$
$1=k\:\sqrt[]{0+4+4+4}$
$1=k\:\sqrt[]{12}$
$k=\frac{1}{\sqrt[]{12}}$
so the unit vector in the direction opposite to w=[0,2,-2,2] is
$ \frac{1}{\sqrt[]{12}}\:\left[0,-2,2,-2 \right] $

Hope this helps

thanks so much!

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