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Helping my child out with their year 11 exam preparation, specifically vectors and dot products, I think I may have figured out the answer but I'd like to get some confirmation or, more likely, a short sharp shock of education :-)

Keep in mind it's some thirty-plus years since I've had to tackle this stuff. The question is phrased thus:

If vector a is perpendicular to vector b-a, which of the following are necessarily true?

1) a.(b-a) = 0
2) a.b = a.a
3) a = b
4) a.b = |a|2

The ones they stated as necessarily true were all but 3.

So here is my reasoning. Consider the vectors as follows. If b-a is perpendicular, then the b vector must be like this (although the triangle could of course be oriented in other ways):

      /|
     / |
    /  |
 b /   | b-a
  /    |
 /     |
/______|
   a

Now, obviously, item 1 is true because the dot product is |a||b-a|cosθ, where θ = 90 hence cosθ = 0.

In terms of the other three statements, I used Pythagoras on the magnitudes to work out:

$${b^2} = {a^2} + {(b-a)^2}$$ $${b^2} = {a^2} + {b^2 -2ab + a^2}$$ $${b^2} = {2a^2} + {b^2 -2ab}$$ $${2a^2} = {2ab}$$ $${a} = {b}$$

So it appears the magnitude of a and b is the same. That would, of course, mean the triangle is not so much a triangle as two congruent lines. This would explain why items 2 and 4 were true - b-a becomes a zero-length vector which I suppose could be considered perpendicular to a.

But the only reason why I can think that item 3 could be false is if that zero-length vector may corrupt things. The other three statements deal with magnitudes only but it may be that that a zero-length vector may be rewritten as zero units north or zero units west, and they may be considered different.

Other than that, I'm not sure why item 3 would not be true as well. Of course, it's quite possible that I've made some mistake in the reasoning above, in which case I'd appreciate some guidance so I can once again become a hero to my son :-)

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    $\begingroup$ Actually, now that I've asked this, I think I realised that treating the b-a vector as having a magnitude of |b|-|a| was wrong - I think that stuffed things up. Still, I'll leave the question here so someone can point out the issue and hopefully let me know the correct way to solve it. $\endgroup$ – paxdiablo Jun 2 at 13:42
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    $\begingroup$ For part $3$: Consider an example. In $\mathbb R^2$, the vector $\vec a=(1,0)$ is orthogonal to $\vec v=(0,3)$, say. Now, to get that in the right form we need $\vec b$ with $\vec v=\vec b-\vec a\implies \vec b=\vec v+\vec a=(1,3)$. We note that $\vec a\neq \vec b$ in this case. $\endgroup$ – lulu Jun 2 at 13:43
  • $\begingroup$ The question asks 'necessarily true', right? If $a = 0$, then is the condition necessarily true? I hope that answers your question! $\endgroup$ – Firefox1921 Jun 2 at 13:46
  • $\begingroup$ @Firefox1921, it was the b-a vector that was zero length, not the a. My (almost certainly faulty) reasoning with Pythagoras indicated that this was required to be the case. $\endgroup$ – paxdiablo Jun 2 at 14:03
  • $\begingroup$ I see that now @paxdiablo. I stand corrected :) $\endgroup$ – Firefox1921 Jun 2 at 16:53
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I don't know what your son is supposed to know about the dot product... which is critical for the answer.

However if he knows that dot product is distributive vs. addition then $$a \cdot (b-a) = a \cdot b - a \cdot a=0.$$

Therefore 2. is clear and 4. also as $a \cdot a = \vert a \vert^2$.

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    $\begingroup$ He says he didn't know it was distributive over addition but, given he's a teenager, it probably just fell out of his head on the way home one day :-) This certainly makes things a hell of a lot easier than my (mis-)method. Thanks heaps. $\endgroup$ – paxdiablo Jun 2 at 14:05
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    $\begingroup$ I'm French and we'll try to remember the expression just fell out of his head on the way home one day that I like! $\endgroup$ – mathcounterexamples.net Jun 2 at 14:09
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Your figure is an example of how $a\cdot(b - a)$ could be true. Note that $b$ is a different vector from $a$ in that figure, so it is an example where $a = b$ is false.

For the rest, in general, if vectors $x$ and $y$ are perpendicular the $x \cdot y = 0.$ This is often used as part of the definition of what it means for vectors to be perpendicular.

But $a\cdot (b - a) = a\cdot b - a \cdot a$ (distribution of dot product over addition), so $a\cdot (b - a) = 0$ implies $a\cdot b - a \cdot a = 0$, which implies $a\cdot b = a \cdot a.$ Moreover, in general for any vector $a,$ we always have $|a|^2 = a\cdot a.$

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  • $\begingroup$ Thanks, David. I knew about the dot product being zero for perpendicular vectors, I'd just forgotten long ago about distribution. Appreciate the help. $\endgroup$ – paxdiablo Jun 2 at 14:07

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