# Two vector problems I need some help with

• Aug 19th 2009, 12:23 PM
Jph93
Two vector problems I need some help with
1. Determine whether the vectors [2,-1] and [-2,-5] are orthogonal.

I remember learning this stuff last year and I think it has something to do with the dot product of vectors maybe? Help would be greatly appreciated.

2. Find the angle between vector u = [-2, 5] and vector v= [-1, 3].

Once again, I completely forget how to do this. Cross product?
• Aug 19th 2009, 12:32 PM
Plato
Quote:

Originally Posted by Jph93
1. Determine whether the vectors [2,-1] and [-2,-5] are orthogonal.
2. Find the angle between vector u = [-2, 5] and vector v= [-1, 3].

If the dot product is zero, $\displaystyle [2,-1]\cdot [-2,-5]=?$ they are orthogonal.

The angle between $\displaystyle u~\&~v$ is $\displaystyle \arccos \left( {\frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}}} \right)$
• Aug 19th 2009, 12:52 PM
Jph93
So -4+5= 1
not orthogonal

and

17/(5.39)(3.16)
= 17/17
=1
arccos1=0

so 0 degrees?
• Aug 19th 2009, 01:06 PM
Plato
Quote:

Originally Posted by Jph93
17/(5.39)(3.16)
= 17/17
=1
arccos1=0

You have missed the lengths.
$\displaystyle \left\| u \right\| = \sqrt {29} \;\& \,\left\| v \right\| = \sqrt {10}$
• Aug 19th 2009, 01:25 PM
Jph93
They equal 17 when multiplied right?
• Aug 19th 2009, 01:51 PM
Plato
Quote:

Originally Posted by Jph93
They equal 17 when multiplied right?

Oh NO!
$\displaystyle \sqrt {10} \cdot \;\sqrt {29} \ne 17\;,\,17^2 = 289 \ne 290$
• Aug 19th 2009, 02:58 PM
QM deFuturo
Quote:

Originally Posted by Plato
If the dot product is zero, $\displaystyle [2,-1]\cdot [-2,-5]=?$ they are orthogonal.

The angle between $\displaystyle u~\&~v$ is $\displaystyle \arccos \left( {\frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}}} \right)$

I am still re-learning my Algebra and have not started on pre-calc yet, but from what I remember about vectors, this notation [2,-1] is the distance of the vector from the origin in the x and y directions, correct?

If this is the case, could you not also solve this problem by calculating then comparing the slope of both vectors as lines?

ie, $\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}, \,$

$\displaystyle m_1 = \frac{-1 - 0}{2 - 0} = \frac{-1}{2}, \,$

$\displaystyle m_2 = \frac{-5 - 0}{-2 - 0} = \frac{5}{2}$

since $\displaystyle m_1 \neq \frac{-1}{m_2}$, vectors are not orthogonal.

Is this approach correct?
• Aug 19th 2009, 03:07 PM
Plato
Well, that is correct as far as it goes.
But notice that the OP is about vectors, whereas you have used traditional analytical geometry.
• Aug 19th 2009, 03:57 PM
dwat
Quote:

Originally Posted by Plato
If the dot product is zero, $\displaystyle [2,-1]\cdot [-2,-5]=?$ they are orthogonal.

The angle between $\displaystyle u~\&~v$ is $\displaystyle \arccos \left( {\frac{{u \cdot v}}{{\left\| u \right\|\left\| v \right\|}}} \right)$

First one not orthogonal because the dot product is not 0?

So Plato, is the answer for the angle one $\displaystyle 0.05875$ radians or $\displaystyle 3.37$ degrees?

Ohh I know this ain't my problem but I stumbled across it and found it good practise on recapping my knowledge on these things...
• Aug 19th 2009, 04:30 PM
Plato
Quote:

Originally Posted by dwat
, is the answer for the angle one $\displaystyle 0.05875$ radians or $\displaystyle 3.37$ degrees?

That is approx correct.
If the angle between two vectors is $\displaystyle 0\text{ or }\pi$ then the vectors must be multiples of one another.
That is clearly not the case here.