I'm just working on some summer problems so that I can be more prepared when I go into my class in the fall. I found a website full of problems of the content we will be learning but it doesn't have the answers. I need a little guidance on how to do this problem. Here is the problem:
Consider the lines $L1$ and $L2$ with equations:
$L_1 : r = (11, 8, 2) + s(4, 3, -1)$
$L_2 : r = (1, 1,-7) + t(2, 1, 11)$
The lines intersect at point $P$.
a. Find the coordinates of $P$.
Would we use a formula for this? For example: $\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}$.
b. Show that the lines are perpendicular.
Isn’t there also a formula to find if lines are perpendicular?
Saying that $L_1:r=(11,8,2)+s(4,3,−1)$ means that every point on the line is of the form (x, y, z)= (11+ 4s, 8+ 3s, 2- s) for some number s. In other words, x= 11+ 4s, y= 8+ 3s, z= 2- s.
$L_2:r=(1,1,−7)+t(2,1,11)$ means that every point on the line is of the form (x, y, z)= (1+ 2t, 1+ t, -7+ 11t) for some number t. In other words, x= 1+ 2t, y= 1+ t, z= -7+ 11t.
Where the two lines intercept the x, y, z coordinates must be the same: 11+ 4s= 1+ 2t, 8+ 3s= 1+ t, 2- s= -7+ 11t.
Notice that these are three equations in only two unknowns, s and t. In general, two lines in three dimensions do NOT intersect- most often they are "skew". What we can do is solve two of the equations for s and t then see if those s and t satisfy the third equation. 11+ 4s= 1+ 2t is the same as 10+ 4s= 2t or 5+ 2s= t. Setting t= 5+ 2s in the second equation, 8+ 3s= 1+ t= 6+ 2s. s= -2 and then t= 5- 4= 1.
Checking the third equation, 2- s= -7+ 11t, 2- (-2)= 4= -7+ 11. Yes, those values are the same so the lines intersect at x= 11+ 4(-2)= 11- 8= 3, y= 8+ 3s= 8+ 3(-2)= 8- 6= 2, z= 2- s= 2- (-2)= 4. The two lines intersect at P= (3, 2, 4).