1. ## Simple 3D Vector geometry question about planes and lines!

How would you find the vector equation of the line h, if told it lies in the plane:

E: x plus y = 4,

and it passes through Q(2,2,1) which is on the plane,

and the line is also perpendicular to the line -->

g = (1 plus µ)i plus (3-µ)j plus (3)k

find line h.

2. Originally Posted by Yehia
How would you find the vector equation of the line h, if told it lies in the plane:

E: x plus y = 4, and it passes through Q(2,2,1) which is on the plane,

and the line is also perpendicular to the line -->

g = (1 plus µ)i plus (3-µ)j plus (3)k

Find line h.
1. Not sure in which part of Germany you live but where I live we would use this notation:

$E:x+y=4$

$g:\vec x = \begin{pmatrix}1\\3\\3\end{pmatrix} + \mu \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix}$

2. You are looking for a point P with the following properties:

$p\in E~\wedge~P\in g~\wedge~(PQ) \perp g$

That means the distance $|\overline{PQ}|$ has a minimum.

There is a simple formula which yields the perpendicular distance of a point to a straight line which is probably known to you. I'll show you a different approach (it is always useful to have an alternative at hand ) :

3. Calculate the vector from Q to any point of g:

$\vec d = \begin{pmatrix}1\\3\\3\end{pmatrix} + \mu \cdot \begin{pmatrix}1\\-1\\0\end{pmatrix} - \begin{pmatrix}2\\2\\1\end{pmatrix}$

$|\vec d| = \sqrt{6-4\mu +2\mu ^2}$

Determine the minimum of $\vec d$ by differentiating wrt $\mu$ and set the numerator of the quotient equal to zero.
You'll get $\mu = 1$ which produces the point P(2,2,3).

4. Thus the line you are looking for is the line $\overline{PQ}$.