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- #1

- Feb 29, 2012

- 342

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.

[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.