Cross product implying vectors are equal For 3-D vectors $a$, $b$ prove:
$a \times b = a − b$ implies $a = b$
I've been working on this question for a while and have no idea how to solve it, any help would be greatly appreciated, thanks.
 A: It is easy to show that $a = 0$ if and only if $b = 0$, so we may assume both are non-zero.
Now, $a \times b = a - b$ implies $(a \times b) \times b = a \times b$, but this is only possible if $a\times b = 0$, so that $a = cb$ for some scalar $c$.
Plugging this back into the original equation, we get $(c - 1)b = 0$. But since $b$ is non-zero by assumption, we get $c = 1$ and so $a = b$.
A: *

*If $a, b$ are linearly independent, then $a\times b$ is orthogonal to
both $a$ and $b$, hence cannot be a linear combination of $a$ and $b$.


*If $a, b$ are linearly dependent that is one is a scalar multiple of
the other, then $a\times b=0$ and $a\times b = a-b\Rightarrow a=b$.
A: If you know that $a \perp (a \times b)$ and $b \perp (a \times b)$, then
\begin{align*}
  \|a \times b\|^2
  &=
  (a \times b) \cdot (a \times b)
  \\
  &=
  (a - b) \cdot (a \times b)
  \\
  &=
  a \cdot (a \times b)
  -
  b \cdot (a \times b)
  \\
  &=
  0 + 0 = 0.
\end{align*}
That is, $a - b = a \times b = \vec{0}$.
And therefore, $a = b$.

In general, if $c \perp a$ and $c \perp b$, then $c$ is perpendicular to the whole space spawned by $a$ and $b$ (vectors of the form $\alpha a + \beta b$). In this case, this means $(a \times b) \perp (a \times b)$.
A: $a\times b$ is a vector which is perpendicular to the plane containing $a$ and $b$. However, a linear combination of $a$ and $b$, i.e. $ma+nb$, is a vector which lies on the plane of $a$ and $b$. This simply implies that the only possible case is that both vectors, i.e. $a\times b$ and $a - b$,are null. Hence, $a-b = 0 \implies \vec a=\vec b$.
A: The cross-product of two vectors is orthogonal to those vectors.
$(a\times b) = a-b\\
(a-b)\cdot(a\times b) = (a-b)\cdot(a-b)\\
a\cdot(a\times b) - b\cdot(a\times b) = \|a-b\|^2\\
0-0 = \|a-b\|^2\\
a-b = 0\\
a=b$
A: The three-dimensional cross product of two vectors, $a \times b$, is always orthogonal to both of its vectors, in other words perpendicular to them, so it is never equal to $a$ or $b$ multiplied by any scalar except zero.
In other words, if you have an equation
$$a\times b = ia+jb$$
where $i$ and $j$ are scalars, the only possible answers are either or both of the conditions $i=j=0$, or $i=-j$ and $a=b$, since the right side of this equation must equal zero.
Just adding my opinion, but the way this problem is phrased seems completely mad. Surely it should say "The condition $a\times b=a-b$ is only satisfied if $a=b$." The whole "implies" part of the question seems to be deliberately perverse phrasing.
