# Analytic Geometry in Space

Can someone help me solve the following two questions:

1) Find the distance between the lines: $$L_1: \frac{x-1}{2} = \frac{y+3}{1} = \frac{z}{-1}$$ and $$\displaystyle L_2 : \frac{x+2}{-2}=\frac{y+5}{3} = \frac{z-1}{-5}$$ I've tried taking an arbitrary point on $L_1$ and build the line passing through it and prependcular to $L_1$ , but couldn't figure out how to do it.

2) Find the equation of the plane passing through $(2,-1,5)$ and perpendicular to the planes $3x-2y+z+7=0$ and $5x-4y+3z+1=0$.

Thanks in advance !

-
For 1: you can use the formula here, after a few suitable conversions... –  Guess who it is. May 7 '12 at 14:45
For 2: the line of intersection of the planes $P_1:\ 3x-2y+z+7=0$ and $P_2:\ 5x-4y+3z+1=0$ is normal to the plane, $P_3$, that you are trying to find. A direction vector of this line is your normal vector to the plane $P_3$. To get this vector, you may take the cross product of the normal vectors of $P_1$ and $P_2$. –  David Mitra May 7 '12 at 15:21

For (1), I follow @DonAntonio in describing the two lines as $P_1+sv_1=f(s)$ and $P_2+tv_2=g(t)$, where $P_1=(1,-3,0)$, $v_1=(2,1,-1)$, $P_2=(-2,-5-1)$, and $v_2=(-2,3,-5)$. But from here on we part ways, because finding the minimal distance between $f(s)$ and $g(t)$ is intrinsically a two-variable problem: we need to find $s$ and $t$ so that the vector $g(t)-f(s)$ is simultaneously perpendicular to $v_1$ and $v_2$.

This is “just” a pair of linear equations in $s$ and $t$, but not such an easy one because the numbers are not nice. The orthogonality relations are $$(P_2-P_1-sv_1+tv_2)\bullet v_j=0\,,$$ for $j=1,2$, in other words we have the two equations $-V_{11}s+V_{12}t=\Pi\bullet v_1$ and $-V_{12}s+V_{22}t=\Pi\bullet v_2$, if we call $V_{ij}=v_i\bullet v_j$ and $\Pi=P_1-P_2$.

Now, what are the numbers? $V_{11}=6$, $V_{12}=4$, and $V_{22}=38$, while $\Pi\bullet v_1=9$ and $\Pi\bullet v_2=5$. This means that the equations to solve are $-6s+4t=9$ and $-4s+38t=5$.

Well, my arithmetic can’t be relied on, but I get $s=-161/106$ and $t=-3/106$; you plug these in to get a hope that $Q_1=f(-161/106)$ and $Q_2=g(-3/106)$ are the two points closest to each other, each on its respective line.

-

Some ideas: for (1) write the lines vector-parametrically equating all the fractions to a real parameter $\,t\,$: $$L_1:x=1+2t\,\,,\,\,y=-3+t\,\,,\,\,z=-t\Longrightarrow L_1: (1,-3,0)+t(2,1,-1)$$

$$L_2: x=-2-2t\,\,,\,\,y=-5+3t\,\,,\,\,z=1-5t\Longrightarrow L_2:\,(-2,-5,1)+t(-2,3,-5)$$

Take now any point in $\,L_1\,$ , say $\,(1,-3,0)\,$ and find the point in $\,w\in L_2\,$ for which $\,\overrightarrow{(1,-3,0)w}\perp(2,1,-1)\,$: $$w\in L_2\Longrightarrow w=(-2t-2\,,\,3t-5\,,\,-5t+1)\Longrightarrow \overrightarrow{(1,-3,0)w}=(-2t-3\,,\,3t-2\,,\,-5t+1)$$ and then we get $$(-2t-3\,,\,3t-2\,,\,-5t+1)\perp(2,1-1)\Longrightarrow 0=-4t-6+3t-2+5t-1\Longrightarrow$$ $$\Longrightarrow 4t=9\Longrightarrow t=\frac{9}{4}$$ Now substitute and calculate with the usual formular the distance between the two point. Disclaimer: The above is a lengthy process and there are pre-calcualted formulae. The above is just one way to do it. Another one uses differential calculus, etc.

For two: you need a plane with a normal $\,(a,b,c)\,$ perpendicular to both given planes' normals , i.e. s.t. $$(i)\,\,(a,b,c)\cdot (3,-2,1)=0\Longrightarrow\,\,I\,\, 3a-2b+c=0$$ $$(ii)\,\,(a,b,c)\cdot(5,-4,3)=0\Longrightarrow \,\,II\,\,5a-4b+3c=0$$

Now solve this system, say $$\,2I-II=a-c=0\Longrightarrow a=c\Longrightarrow \,\,I\,\,\,3a-2b+a=0\Longrightarrow b =2a$$

So the wanted plane has as normal any vector of the form $\, (a,2a,a)\,\,,\,\,a\neq 0\,$ , and as we want it to pass throught the point $\,(2,-1,5)\,$ we then get $$2a-2a+5a+d =0\Longrightarrow d=-5a$$so choosing, say $\,a=1\,$ we get the plane $$x+2y+z-5=0$$

-