
Lines in a Plane 13
Find a vector equation for the line through the origin that intersects both of the lines r=(2,16,19)+t(1,1,4) and r=(14,19,2)+u(2,1,2).
This is my work so far, but I can't seem to find the correct answer.
d1=(1,1,4)
d2=(2,1,2)
i'm stuck here..

Let r=(0,0,0)+v(a,b,c) the requested line
There exists a value for v, say $\displaystyle v_1$ and another value, say $\displaystyle v_2$ such that
$\displaystyle 2+t=v_1a$
$\displaystyle 16+t=v_1b$
$\displaystyle 194t=v_1c$
$\displaystyle 142u=v_2a$
$\displaystyle 19+u=v_2b$
$\displaystyle 2+2u=v_2c$
Let $\displaystyle x = \frac{v_2}{v_1}$
$\displaystyle 142u=x(2+t)$
$\displaystyle 19+u=x(16+t)$
$\displaystyle 2+2u=x(194t)$
Solving gives
$\displaystyle x = \frac{64}{9}$
$\displaystyle u = 41$
$\displaystyle t = \frac{121}{16}$
You can chose one of the coordinates a, b or c
If you chose a = 17 then b = 15 and c = 20


I can't seem to figure out how you got your final answer, the a=17 then b=15 and c =20.. this step before this, or to get to this a=17, where did it come from?
Can you, runninggag, or someone else clarify it for me? please and thank you

As I said before you can chose any value for a
I have chosen a=17 because it makes the other values very simple
The coordinates of the intersect with r=(2,16,19)+t(1,1,4) is given for t=121/16, therefore (153/16,135/16,180/16)
If you chose a=17, you will get b=15, c=20 and v1=9/16
But if you chose a=34, you will get b=30, c=40 and v1=9/32
This does not change the coordinates of the intersect (v1a,v1b,v1c) and does not change either the line since r=(0,0,0)+v(17,15,20) and r=(0,0,0)+v(34,30,40) are the same line

ok, so i could even choose a=1 and then find v1, and then find b and c right?
so
a=1
v1=153/16
b=135/153
c=180/153

runninggag? can you let me know about what i said above? just this last question
Thanks so much for your help by the way :)

That's correct ! (Clapping)