# Vectors in 3D question

• Apr 5th 2010, 12:40 PM
kandyfloss
Vectors in 3D question
lines L1 and L2 have vector equations r=8i-j+3k+ λ(-4i +j)
and r= -2i +8j - k +
µ(i +3j - 2k) respectively. it then says show that L1 and L2 intersect and find the position vector of the point of intersection.
• Apr 5th 2010, 02:23 PM
Soroban
Hello, kandyfloss!

Quote:

Lines $\displaystyle L_1$ and $\displaystyle L_2$ have vector equations: .$\displaystyle \begin{Bmatrix}\vec r&=&8\vec i-\vec j+3\vec k+ \lambda\left(-4\vec i +\vec j\right) \\ \\[-3mm] \vec r &=& \text{-}2\vec i +8\vec j - \vec k + \mu\left(\vec i +3\vec j - 2\vec k\right) \end{Bmatrix}$

Show that $\displaystyle L_1$ and $\displaystyle L_2$ intersect and find the position vector of the point of intersection.

The vector equations can be written:

.$\displaystyle L_1\!:\;\;\begin{Bmatrix}x &=& 8 - 4\lambda & [1] \\ y &=& \text{-}1 + \lambda & [2] \\ z &=& 3 & [3] \end{Bmatrix}$ . . . $\displaystyle L_2\!:\;\;\begin{Bmatrix}x &=& \text{-}2 + \mu & [4] \\ y &=& 8 + 3\mu & [5] \\ z &=& \text{-}1 - 2\mu & [6] \end{Bmatrix}$

$\displaystyle \begin{array}{ccccccccc} \text{Equate [1] and [4]:} & 8-4\lambda &=& \text{-}2 + \mu & \Rightarrow & 4\lambda + \mu &=& 10 & [7] \\ \text{Equate [2] and [5]:} & \text{-}1 + \lambda &=& 8 + 3\mu & \Rightarrow & \lambda - 3\mu &=& 9 & [8] \\ \text{Equate [3] and [6]:} & 3 &=& \text{-}1 - 2\mu & \Rightarrow & \text{-}2\mu &=& 4 & [9]\end{array}$

From [9], we have: .$\displaystyle {\color{blue}\mu \:=\:-2}$

Substitute into [8]: .$\displaystyle \lambda - 3(\text{-}2) \:=\:9 \quad\Rightarrow\quad {\color{blue}\lambda \:=\:3}$

Substitute $\displaystyle \lambda = 3$ into [1], [2] and [3]:

. . $\displaystyle L_1\!:\;\;\begin{Bmatrix}x &=& 8-4(3) &=& \text{-}4 \\ y &=& \text{-}1 + 3 &=& 2 \\ z &=& 3 &=& 3\end{Bmatrix}$

Substitute $\displaystyle \mu = -2$ into [4], [5] and [6]:

. . $\displaystyle L_2\!:\;\;\begin{Bmatrix}x &=& \text{-}2 - 2 &=& \text{-}4 \\ y &=& 2 + 3(-2) &=& 2 \\ z &=& \text{-}1 - 2(\text{-}2) &=& 3 \end{Bmatrix}$

Therefore, the lines intersect at: .$\displaystyle P(\text{-}4,2,3)$

The position vector of $\displaystyle P$ is: .$\displaystyle \vec p \:=\:\langle \text{-}4,2,3\rangle \;=\;\text{-}4\vec i + 2\vec j + 3\vec k$

• Apr 12th 2010, 10:48 AM
kandyfloss
Excellent explanation!
thanks (Happy)