Well, we've learned how to detect whether two vectors are perpendicular to each other using dot product.
a.b=0
if two vectors parallel, which command is relatively simple.
for 3d vector, we can use cross product.
for 2d vector, use what?
for example,
a={1,3}, b={4,x};
a//b
How to use a equation to solve $x$.
I tried it, but this is a little complex.
Projection[a, b] - a == 0
$a$ and $b$ are parallel if $a = \kappa b$. Try
MatrixRank[{a, b}] == 1
for an easy way to test this. This works only if neither of the vectors have norm 0. Symbolic vector components (parameters) are considered independent by MatrixRank, so this method considers vectors parallel only if they are parallel for any value of the parameters.
For a fully general symbolic solution use
Reduce[a = k b, k]
a = {x, y};
b = {x, z};
Reduce[a == k b, k]
(* (y == z && k == 1) || (x == 0 && z != 0 && k == y/z) || (z == 0 && y == 0 && x == 0) *)
a = {1, 3};
b = {4, x};
Reduce[a == k b, {k, x}]
(* k == 1/4 && x == 12 *)
a={1,2}, b={3,x},a//b, I hope solve a equation to get value of $x$. how to do this?
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Commented
Nov 29, 2016 at 12:15
Reduce .
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a = {1, 2};b = {4, x};Reduce[a == k *b && k \[Element] Integers, {x, k}] this do not work.
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Commented
Nov 29, 2016 at 12:26
k==1/4, which is not an integer.
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Based on the definition of the scalar product for $d$-dimensional vectos
\begin{equation} a \cdot b = |a| |b| \cos(\phi) \end{equation}
you can create a test based only on vector operations. The vectors are parallel, if and only if the angle between them is $\phi = 0$ or $\phi = \pi$. So, if the following quantity $q$
\begin{equation} q = \frac{|a \cdot b|}{|a| |b|} \end{equation}
equals to 1, then the vectors are parallel. Mathematica code
(*Dimension*)
d = 7;
(*Generate vectors*)
a = RandomReal[{-1, 1}, d];
b = RandomReal[]*a;
(*Test*)
If[Abs[a.b]/(Norm[a]*Norm[b]) == 1, Print["Parallel"],
Print["Not parallel"]]
Parallel
Based on $q$ you can solve your problem for $x$. But I think you can figure that out.
Reduce[(a.b)^2 == a.a b.b] is indeed better than my suggestion of Reduce[a==k b]. For a={x,y}; b={x,z} your way gives the very clean x == 0 || y == z. Probably this should be the accepted answer.
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Try this:
rule = {x_, y_} -> {x, y, 0};
Cross[a /. rule, b /. rule]
if
a = {1, 1};
b = {0.1, 0.1};
Cross[a /. rule, b /. rule]
(* {0., 0., 0.} *)
or alternatively
Cross @@ Map[Replace[#, rule] &, {a, b}]
Have fun!
May be it's simple:
Det@{a,b}==0
