lines in 3 dimensional space

• May 26th 2008, 12:45 PM
imthatgirl
lines in 3 dimensional space
i have 2 questions here which probably aren't as hard as i think but my teacher never taught me how to do it, so can anyone help?

1) determine if the vector [1,-1,5] is perpendicular to [x,y,z] = [3,-1,5] + t [2,-3,-1]

2) Determine if these two equations intersect
[x,y,z] = [4,-3,2] + t [1,8,-3]
[x,y,z] = [2,-19,8] + t [4,-5,-9]

...actually i got it.. i figured it out..thanks anyways
• May 26th 2008, 04:24 PM
Soroban
Hello, imthatgirl!

Quote:

1) Determine if the vector $\vec{u} \:=\:[1,\text{-}1,5]$ is perpendicular
to the line: . $\vec{L} \:=\:[3,\text{-}1,5] + t[2,\text{-}3,\text{-}1]$

Two vectors are perpendicular if their dot product is zero.

We have vector $\vec{u} \:=\:[1,\text{-}1,5]$
The line has vector: . $\vec{v} \:=\:[2,\text{-}3,\text{-}1]$

Then: . $u\cdot v \:=\:[1,\text{-}1,5]\cdot[2,\text{-}3,\text{-}1] \:=\:(1)(2) + (\text{-}1)(\text{-}3) + (5)(\text{-}1) \:=\:2 + 3 - 5 \;=\;0$

Therefore: . $\vec{u} \perp \vec{L}$

Quote:

Determine if these two lines intersect.

. . $[x,y,z] \;= \;[4,\text{-}3,2] + t[1,8,\text{-}3]$
. . $[x,y,z] \;=\; [2,\text{-}19,8] + u[4,\text{-}5,\text{-}9]$

We have: . $\begin{Bmatrix}x & = & 4 + t \\ y &=&\text{-}3 + 8t \\ z &=& 2 - 3t\end{Bmatrix} \quad \begin{Bmatrix}x &=& 2 + 4u \\ y &=& \text{-}19 - 5u \\ z &=& 8 - 9u \end{Bmatrix}$ .(A)

Equate components: . $\begin{array}{cccccccc}4 + t &=& 2 + 4u & \Rightarrow & t - 4u &=& \text{-}2 &{\color{blue}[1]} \\ \text{-}3 + 8t &=& \text{-}19 - 5u & \Rightarrow & 8t + 5u &=&\text{-}16 & {\color{blue}[2]}\\ 2 - 3t &=& 8-9u & \Rightarrow & t - 3u &=& \text{-}2 & {\color{blue}[3]}\end{array}$

Subtract [1] from [3]: . $u \:=\:0$
Substitute into [1]: . $t \:=\:-2$

Substitute into (A) and we get: . $(x,y,x) \;=\;(2,-19,8)$ . . . intersection!