1. ## Vector question 2-

The line L passes through the point A, whose position vector is i-j-5k and is parrallel to the vector i-j-4k. The line l2 passes through the point B, whose position vector is 2i-9j-14k, and is parrallel to the vector 2i+5j+6k. The point L and the point Q on L2 are such that PQ is perpendicular to both L and L2.

a. Find the length of PQ
b. Find a vector perpendicular to the plane which contains PQ and L2.
c. Find the perpendicular distance from A to the plane.

2. ## part a) only

Originally Posted by kingkaisai2
The line L passes through the point A, whose position vector is i-j-5k and is parrallel to the vector i-j-4k. The line l2 passes through the point B, whose position vector is 2i-9j-14k, and is parrallel to the vector 2i+5j+6k. The point L and the point Q on L2 are such that PQ is perpendicular to both L and L2.

a. Find the length of PQ...
Hi, have a look at the attached image.

I used the writing of vectors I'm used to. So maybe you have to transscript the vectors into the form you should use:

3. If $\displaystyle l_1 :A + tD\quad \& \quad l_2 :B + tE$ are two skew lines, then the minimum distance between them is $\displaystyle \frac{{\left| {\left( {B - A} \right) \cdot \left( {D \times E} \right)} \right|}}{{\left\| {D \times E} \right\|}}.$ That is the length of the PQ in your question.

The plane that contains $\displaystyle l_2$ and PQ is $\displaystyle \left[ {D \times \left( {D \times E} \right)} \right] \cdot \left( { < x,y,z > - D} \right) = 0.$

4. ## part b) only

Originally Posted by kingkaisai2
...
b. Find a vector perpendicular to the plane which contains PQ and L2.
...
I've attached an image with my calculations:

5. Hello, kingkaisai2!

Here's my approach to part (a).
. . But I'm sure there's a better way!

The line $\displaystyle L_1$ passes through the point $\displaystyle A(1,\text{-}1,\text{-}5)$
. . and is parallel to the vector $\displaystyle \vec{u} \:=\:\langle1,\text{-}1,\text{-}4\rangle$
The line $\displaystyle L_2$ passes through the point $\displaystyle B(2,\text{-}9,\text{-}14)$
. . and is parallel to the vector $\displaystyle \vec{v}\:=\:\langle 2,5,6\rangle.$
The point $\displaystyle P$ on $\displaystyle L_1$ and the point $\displaystyle Q$ on $\displaystyle L_2$ are such that
. . $\displaystyle PQ \perp L_1$ and $\displaystyle PQ \perp L_2.$

(a) Find the length of $\displaystyle PQ$

Line $\displaystyle L_1$ has parametric equations; .$\displaystyle \begin{Bmatrix}x \,= \,1 + t \\ y \,= \,\text{-}1 - t \\ z \,= \,\text{-}5 - 4t\end{Bmatrix}$

Line $\displaystyle L_2$ has parametric equations: .$\displaystyle \begin{Bmatrix}x\,=\,2 + 2u \\ y \,=\,\text{-}9 + 5u \\ z\,=\,\text{-}14 + 6y\end{Bmatrix}$

The vector perpendicular to both $\displaystyle \vec{u}$ and $\displaystyle \vec{v}$ is given by:
. . $\displaystyle \vec{n}\;=\;\begin{vmatrix}i & j & k \\ 1 & \text{-}1 & \text{-}4 \\ 2 & 5 & 6\end{vmatrix} \;=\;\langle14,-14,7\rangle \;=\;\langle2,-2,1\rangle$

We have: .$\displaystyle P(1+t,\,\text{-}1-t,\,\text{-}5,-4t)$ and $\displaystyle Q(2+2u,\,\text{-}9+5u,\,\text{-}14+6u)$

. . and we want: .$\displaystyle \overrightarrow{PQ} \parallel \vec{n}$

So we have: .$\displaystyle \overrightarrow{PQ} \;=\;\langle\,(2\!+\!2u) - (1\!+\!t),\;(\text{-}9\!+\!5u) - (\text{-}1\!-\!t),\;(\text{-}14\!+\! 6u) -(\text{-}5 - 4t)\,\rangle$
. . . . . . . . . .$\displaystyle PQ\;=\;\langle 2u - t + 1,\;5u + t- 8,\;6u + 4t- 9\rangle$

$\displaystyle \text{If }\overrightarrow{PQ} \parallel \vec{n}\text{, then: }\;\langle 2u-t+ 1,\:5u+t-8,\:6u + 4t=9\rangle \:=\:k\langle2,-2,1\rangle$

And we have: .$\displaystyle \begin{Bmatrix}2u - t +1= 2k \\ 5u + t -8= \text{-}2k \\ 6u + 4t -9 = k\end{Bmatrix} \;\Rightarrow\;\begin{Bmatrix}2u - t - 2k = \text{-}1 \\ 5u + t + 2k = 8 \\ 6u + 4t - k = 9\end{Bmatrix}$

Solve the system of equations: .$\displaystyle k = 1,\:t = 1,\:u = 1$

Hence: .$\displaystyle P(2,-2,-9)$ and $\displaystyle Q(4,-4,-8)$

Therefore: .$\displaystyle \overline{PQ}\;=\;\sqrt{(4-2)^2 + (-4+2)^2 + (-8+9)^2} \;=\;\sqrt{4 + 4 + 1} \;=\;\sqrt{9}\;=\;\boxed{3}$