# Finding a linear equation

I cannot understand this question:

"Find a linear equation (and parametrics) to $v$ where $v$ is perpendicular to the line segment of the extremes $(1,2,1)$ and $B$ $(1,8, -5)$, dividing it in half."

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That is a strange way to phrase the question. My best guess is that it asks you to parametrize the plane which is (a) perpendicular to the line segment and (b) cuts it in half. – us2012 Mar 17 '13 at 19:39

The question seems to ask you to find the equation of the plane through the midpoint $(1,5,-2)$ of $(1,2,1)$ and $(1,8,-5)$ and which has $(1,2,1)-(1,8,-5)=(0,-6,6)$ as a normal. Hence it is given by $\vec r\cdot(0,-1,1)=(1,5,-2)\cdot(0,-1,1)$ or $y-z=7$.

Parametrically, this is given by

$\{(x,z+7,z):x,z\in\mathbb R\}$

$=\{(0,7,0)+(x,z,z):x,z\in\mathbb R\}$

$=\{(0,7,0)+\lambda(1,0,0)+\mu(0,1,1):\lambda,\mu\in\mathbb R\}$

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+1 Do you need me to attach a diagram for illustrating the vectors $\vec{r}\perp \vec{n}$? – kiss my armpit Mar 17 '13 at 20:21

The midpoint of segment $AB$ is $A*B=(1,5,-2)$ and $(a,b,c)^T=(0,6,-6)^T$ is a direction vector to the line $AB$ so the desired equation of the plane is $$ax+by+cz=d,$$ where $d$ is determined by substituting the coordinates of the midpoint in the equation.

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@HenningMakholm No $(0,6,-6)$ is a direction vector of the line $AB$. – user63181 Mar 17 '13 at 19:53