# Analytic Geometry in Space

#### Fantini

MHB Math Helper
I'm trying to help a friend but I don't remember any of this, so it'd help us both. Book recommendations are also welcomed.

Given the vector equations $$r: \begin{cases} x = 2 - \lambda, \\ y = 1 + 3 \lambda, \\ z = 1 + \lambda, \end{cases} \text{ and } s: \begin{cases} x = 1+t, \\ y = 3+4t, \\ z = 1 + 3t, \end{cases}$$

find:

[a] the equation of the plane that contains the line $s$ and it's parallel to $r$;
the distance of the point $P_0 = (2,0,0)$ to the line $s$;
[c] the distance between the lines $r$ and $s$;
[d] a point $P$ in $r$ and a point $Q$ in $s$ such that the distance between $P$ and $Q$ be equal to the distance between $r$ and $s$.

Don't necessarily need the full solution (although it'd be appreciated), hints and tips on the train of thought will be hugely valuable.

#### earboth

##### Active member
I'm trying to help a friend but I don't remember any of this, so it'd help us both. Book recommendations are also welcomed.

Given the vector equations $$r: \begin{cases} x = 2 - \lambda, \\ y = 1 + 3 \lambda, \\ z = 1 + \lambda, \end{cases} \text{ and } s: \begin{cases} x = 1+t, \\ y = 3+4t, \\ z = 1 + 3t, \end{cases}$$

find:

[a] the equation of the plane that contains the line $s$ and it's parallel to $r$;
the distance of the point $P_0 = (2,0,0)$ to the line $s$;
[c] the distance between the lines $r$ and $s$;
[d] a point $P$ in $r$ and a point $Q$ in $s$ such that the distance between $P$ and $Q$ be equal to the distance between $r$ and $s$.

Don't necessarily need the full solution (although it'd be appreciated), hints and tips on the train of thought will be hugely valuable.

to a)

The equation of the plane must contain the line s completely and the direction vector of r:

$\displaystyle{\langle x,y,z \rangle = \langle 1,3,1 \rangle + t \cdot \langle 1,4,3 \rangle + \lambda \cdot \langle -1,3,1 \rangle}$

That's all.

to b)

Determine the minimum of the distance from $P_0$ to any point of the straight line:

$Q \in s$. Then the distance is:

$d = |\overrightarrow{P_0,Q}| = |\vec q - \overrightarrow{p_0}|$

That means:

$d(t) = \sqrt{(-1+t)^2+(3+4t)^2+(1+3t)^2} = \sqrt{11+28t+26t^2}$

Now differentiate d, solve the equation d'(t) = 0 for t. Plug in this value into the equation of the straight line. You'll get the point Q whose distance to P is at it's minimum. Calculate the distance $\overline{P_0,Q}$.

to c)

Calculate the perpendicular distance of M(2, 1, 1) to the plane of part a). (Why?)

to d)

$\overrightarrow{PQ}$ must be perpendicular to both lines that means:

$\overrightarrow{PQ} \cdot \langle 1,4,3 \rangle = 0 ~\wedge~ \overrightarrow{PQ} \cdot \langle -1,3,1\rangle = 0$

Solve for $\lambda$ and $t$.