Prove 2 3D lines intersect

[ku1005]

Hey guys, not sure how im meant to do the following, and it seems straightforward.Any pointers greatl appreaciated.

"A line through the point P(1,1,1) is paralell to the vector u=(1,2,3). Another line through Q(2,1,0) is paralell to the vector v=(3,8,13). Prove that the two lines intersect and determine their point of intersection."

I came up with the parameters shoudl be joined as follows

1+k = 2+ 3t (x)
1+2k = 1+ 8t (y)
1+3k = 13t (z)

BUT...what i find is that it suggests they are not intersecting since

say for x , we say k = 1+3t and SUB into the "y"


becoming 1+2(1+3t) = 1+8t...........which to me isnt true???


Where is my misunderstanding in this situation???

cheers...appreaciate it!

rhys

 

[soroban]

Hello, Rhys!

A line through the point P(1,1,1) is paralell to the vector u=(1,2,3).
Another line through Q(2,1,0) is paralell to the vector v=(3,8,13).
Prove that the two lines intersect and determine their point of intersection.

I came up with the parameters should be joined as follows:

. . \(\displaystyle \begin{array}{cccc}1\,+\,k & \:=\: & 2\,+\,3t & \;(x) \\
1\,+\,2k & = & 1\,+\,8t & \;(y) \\
1\,+\,3k &=& 13t& \;(z)\end{array}\)


BUT...what i find is that it suggests they are not intersecting since

say, for \(\displaystyle (x)\) , we have: \(\displaystyle \, k\:=\:1\,+\,3t\,\) and SUB into \(\displaystyle (y)\)

becoming: \(\displaystyle \:1\,+\,2(1\,+\,3t)\:=\:1\,+\,8t\;\) . . . which to me isn't true.
. . Of course not . . . It's not supposed to be an identity.

Where is my misunderstanding in this situation?

Solve that equation: \(\displaystyle \:1\,+\,2\,+\,6t\:=\:1\,+\,8t\;\;\Rightarrow\;\;2t\,=\,2\;\;\Rightarrow\;\;\fbox{t\,=\,1}\)

Substitute into \(\displaystyle (x):\;1\,+\,k\:=\:2\,+\,3\cdot1\;\;\Rightarrow\;\;\fbox{k\,=\,4}\)


Check these values in all three equations.

\(\displaystyle \begin{array}{cccc}(x): & 1\,+\,4 &\:=\:&2\,+\,3\cdot1 & \;\Rightarrow\; &5\,=\,5 \\
(y): & 1\,+\,2\cdot4 & = & 1\,+\,8\cdot1 & \Rightarrow & 9\,=\,9 \\
(z): & 1\,+\,3\cdot4 & = & 13\cdot1 & \Rightarrow & 13\,=\,13
\end{array}\) . . . Check! .The lines intersect.

And they intersect at \(\displaystyle (5,\,9,\,13)\)

 

[ku1005]

ahk...cheers...i see now...thanks for the help