# symmetric equation

• Apr 12th 2008, 11:01 AM
fastman390
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
• Apr 12th 2008, 11:59 AM
Mathstud28
Ok
Quote:

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 $\displaystyle x,y,z$ all equal to t to get parametric equaitions...now it should be easy
• Apr 12th 2008, 12:10 PM
earboth
Quote:

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:

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

to #A) Solve the system of simultaneous equations:

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

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

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

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

and therefore the parametric equation of L3 is:

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