Check to see if two vectors are parallel This is a small excercise from Lang's Introduction to Linear Algebra
Two vectors $\overrightarrow{PQ}$ and $\overrightarrow{AB}$ are defined by these following n-tuples:
$P=(1,4), Q=(-3,5), A=(5,7), B=(9,6)$
I know that two vectors are parallel if $B-A=c(P-Q).$ If $c$ is greater than zero, then two vectors point at the same direction, while if c is smaller than zero, then they point at the opposite direction.
However, I fail to see how these two vectors are parallel. Computing $B-A=c(P-Q)$, we have:
$(B-A)=(4,-1)$ and $(P-Q)=(-4,1)$. How can these two vectors parallel?
The answer in the back of the book says they are parallel. 
 A: $(4,-1) = (-1) \ (-4,1)$
$-1$ is a scalar constant less than zero.
A: It is $$\vec{PQ}=[-4,1]$$ and $$\vec{AB}=[4,-1]$$ and it is $$[-4,1]=-[4,-1]$$
A: The confusion comes in from an imprecise or slightly varying definition of parallel.  At least one math professor, Paul Dawkins, PhD, considers opposite directions to be parallel.  
...Two vectors are parallel if they have the same direction or are in exactly opposite directions. Now, recall again the geometric interpretation of scalar multiplication. When we performed scalar multiplication we generated new vectors that were parallel to the original vectors (and each other for that matter).
and he elaborates in example 2a.
These two vectors are parallel since →b = −3→a
http://tutorial.math.lamar.edu/Classes/CalcII/VectorArithmetic.aspx
On a related note, I consider his website to be an excellent source for teaching Calculus and similar courses.
A: Apply the doc product to $\vec{PQ}=(-4,1)$ and $\vec{AB}=(4,-1)$ to evaluate the angle $\theta$ between them,
$$\cos\theta = \frac{\vec{PQ}\cdot\vec{AB}}{|PQ||AB|} = -1$$
which yields $\theta = -180^\circ$, hence parallel but in opposite directions.
A: $c=-1$ is a solution
Also, you can also determine if two vectors are parallel by leveraging the fact that every vector can be expressed as the product of its length and direction, and noting that two vectors are parallel iff one vector is a scalar multiple of the other...
$\vec{PQ} = <-3-1,5-4> = <-4,1>$
$|\vec{PQ}| = \sqrt{(-4)^2 + 1^2} = \sqrt{17}$
Hence, 
$\vec{PQ} = \sqrt{17} <\frac{-4}{\sqrt{17}}, \frac{1}{\sqrt{17}}>$

$\vec{AB} = <9-5,6-7> = <4,-1>$
$|\vec{AB}| = \sqrt{(4)^2 + (-1)^2} = \sqrt{17}$
Hence, 
$\vec{AB}= \sqrt{17} <\frac{4}{\sqrt{17}}, \frac{-1}{\sqrt{17}}> = -\sqrt{17} <\frac{-4}{\sqrt{17}}, \frac{1}{\sqrt{17}}>$

Note that $\vec{PQ}$ and $\vec{AB}$ can be manipulated algebraically so that they have the same direction and differ only by a scalar multiple. Since, one is a scalar multiple of the other, the two are parallel.
