Okay, so I was given a task that looked easy, but when given the solution...need some explanation.
"Two points, A=(2,-1,8) and B=(0,3,-6), where A is the mirror image of B in regard to the plane π. Find the equation of this plane and the coordinates of the intersection between AB and π".
My thought was: A & B must have the same distance to the plane so if I just do: 1/2 (b - a). I thought I was calculating the intersection point here, but apparently the result was -n of the plane, which make sense.
To find the intersection point the solution says = 1/2(ra + rb), (meaning location vector a plus location vector b split in 2).
How can this be correct? Makes no sense to me so I hope someone here can clear things up!
$\endgroup$ 13 Answers
$\begingroup$$b-a$ is the vector from a to b. $\tfrac{1}{2}(b-a)$ is half of this vector. This vector gives you how to move from $a$ to find a point on your plane. Therefore you need to start from $a$ and follow the vector, which gives you the point $a + \tfrac{1}{2}(b-a)$.
$\endgroup$ 1 $\begingroup$I have no time to draw a diagram at the moment, so suppose that we write:
$$\mathbf a = \mathbf p + \mathbf p^\perp, \mathbf b = \mathbf q + \mathbf q^\perp$$
with $\mathbf p, \mathbf q \in \pi$, and $\mathbf p^\perp, \mathbf q^\perp$ perpendicular to $\pi$ (check that this is unique). Then the operation of mirroring in $\pi$ is simply $\mathbf v + \mathbf v^\perp \mapsto \mathbf v - \mathbf v^\perp$ for any $\mathbf v \in \pi, \mathbf v^\perp$ perpendicular to $\pi$.
Since $\mathbf b$ is $\mathbf a$ mirrored in $\pi$, we thus find that $\mathbf q = \mathbf p$ and $\mathbf q^\perp = - \mathbf p^\perp$.
It follows that $\dfrac12(\mathbf a + \mathbf b) = \dfrac 12(2\mathbf p + \mathbf p^\perp - \mathbf p^\perp) = \mathbf p \in \pi$.
What I've effectively done is to choose a basis for $\Bbb R^3$ suitable to the problem. Your answer is almost correct since $\dfrac12(\mathbf a +\mathbf b) = \mathbf a + \dfrac12(\mathbf b - \mathbf a)$; you had omitted to "move" the difference vector to start at $A$.
$\endgroup$ $\begingroup$I don't understand the distinctions you are making among $A$, $\bf A$, $\bf a$, and $\bf ra$, but if $A$ is the mirror image of $B$ in the plane $\pi$, then the intersection between $AB$ and $\pi$ is halfway between $A$ and $B$, and that's $(A+B)/2$.
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