Cross product implying vectors are equal

Question

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.

Answer 1

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.

Answer 2

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 $a = b$.

Answer 3

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, $a = b$.

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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 $(a \times b) \perp (a \times b)$.

Answer 4

• 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 $a\times b = a-b\Rightarrow a=b$.

Answer 5

The cross-product of two vectors is orthogonal to those vectors.

$(a\times b) = a-b\\ (a-b)\cdot(a\times b) = (a-b)\cdot(a-b)\\ a\cdot(a\times b) - b\cdot(a\times b) = \|a-b\|^2\\ 0-0 = \|a-b\|^2\\ a-b = 0\\ a=b$

Answer 6

$a\times b$ 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, $a-b = 0 \implies \vec a=\vec b$.