# Math Help - symmetric equation

1. ## symmetric equation

L1 : X-4 = Y-8 = Z+1 L2 : X-16 = Y-2 = Z+1
2 3 -4 -6 1 2
A) show that L1 and L2 intersected
B) find parametric equation of the line that passes through the point of intersection of L1 and L2 IS perpandicular to both of them

2. ## Ok

Originally Posted by fastman390
L1 : X-4 = Y-8 = Z+1 L2 : X-16 = Y-2 = Z+1
2 3 -4 -6 1 2
A) show that L1 and L2 intersected
B) find parametric equation of the line that passes through the point of intersection of L1 and L2 IS perpandicular to both of them
set $x,y,z$ all equal to t to get parametric equaitions...now it should be easy

3. Originally Posted by fastman390
L1 : X-4 = Y-8 = Z+1 L2 : X-16 = Y-2 = Z+1
2 3 -4 -6 1 2
A) show that L1 and L2 intersected
B) find parametric equation of the line that passes through the point of intersection of L1 and L2 IS perpandicular to both of them
Re-write the equation:

$L_1:\left\{\begin{array}{lcr}x&=&2t+4 \\ y&=&3t+8 \\ z&=&-4t-1\end{array}\right.$ ..... and ..... $L_2:\left\{\begin{array}{lcr}x&=&-6s+16 \\ y&=&s+2 \\ z&=&2s-1\end{array}\right.$

to #A) Solve the system of simultaneous equations:

$\left|\begin{array}{lcr}-6s+16&=&2t+4 \\ s+2&=&3t+8 \\ 2s-1&=&-4t-1\end{array}\right.$ ..... which will yield $s=\frac{12}5~\wedge~t=-\frac65$

The point of intersection is $P\left(\frac85\ ,\ \frac{22}5\ ,\ \frac{19}5\right)$

to #B: The direction vector of the line L3 is the crossproduct of the direction vectors of L1 and L2:

$(2, 3, -4) \times (-6, 1, 2) = (10, 20, 20)$

and therefore the parametric equation of L3 is:

$L_3:\left\{\begin{array}{lcr}x&=&10r+\frac85 \\ \\ y&=&20r+\frac{22}5 \\ \\ z&=&20r+\frac{19}5\end{array}\right.$