# Thread: Vector ( Distane between 2 lines)

1. ## Vector ( Distane between 2 lines)

Question: Two lines, l1 and l2, have equations r = -2i + k + t(i + j + k) and r = i + 5j + 2k + s(i + 2j + 3k) where s and t are constants. Points P and Q lies on the lines l1 and l2 respectively. If PQ is perpendicular to both l1 and l2, find the coordinates of P and Q and hence show that the length of PQ is √6 .

This is how I solve the problem:
l1 : r = (t - 2) i + t j + (t + 1)k
l2 : r = (1 + s )i + (5 +2s) j + (2 + 3s) k

Let b1 = i + j + k
Let b2 = i + 2j + 3k

b1 × b2 = i – 2j + k

Since line b1 × b2 is perpendicular to l1 and l2 , then I try to use the scalar product of (b1 × b2 ) ∙ P and (b1 × b2 ) ∙ Q to get the constants s and t , but I failed, could anyone help me to solve this problem? Thanks.

2. ## Re: Vector ( Distane between 2 lines)

First I don't think that s and t can be constants ,because if it is, there cannot be a line $\displaystyle l_1$ and $\displaystyle l_2$

Now to the problem,

take the position vector of P and Q are respectively, $\displaystyle (t_0-2)\,\mathbf{i}+t_0\,\mathbf{j}+(t+1)\,\mathbf{k}$ and $\displaystyle (1+s_0)\,\mathbf{i}+(5+2s_0)\, \mathbf{j}+(2+3s_0)\, \mathbf{k}$

then

\displaystyle \begin{align*}\overrightarrow{PQ}&=[(1+s_0)-(t_0-2)]\,\mathbf{i}+[(5+2s_0)-t_0]\,\mathbf{j}+[(2+3s_0)-(t_0+1)]\,\mathbf{k}\\ &=(3+s_0-t_0)\,\mathbf{i}+ (5+2s_0-t_0)\,\mathbf{j}+(1+3s_0-t_0)\,\mathbf{k}\\ \end{align*}

take vectors parallel to $\displaystyle l_1$ and $\displaystyle l_2$ respectively $\displaystyle \underline{b_1}$ and $\displaystyle \underline{b_2}$ then,

$\displaystyle \underline{b_1}= \mathbf{i} +\mathbf{j} + \mathbf{k}$
$\displaystyle \underline{b_2}= \mathbf{i} + 2\,\mathbf{j} + 3\,\mathbf{k}$

So let's consider the given fact now, it says the $\displaystyle \overrightarrow{PQ}$ is perpendicular to both $\displaystyle \underline{b_1}$ and $\displaystyle \underline{b_2}$. This means following dot products should be equal to zero.

$\displaystyle \underline{b_1}.\overrightarrow{PQ}=0$

$\displaystyle \underline{b_2}.\overrightarrow{PQ}=0$

from these two equation you'll get two equations involving $\displaystyle t_0,s_0$. Solve them and find the position vector of P and Q

you'll eventually get $\displaystyle s_0=1, t_0=5$

3. ## Re: Vector ( Distane between 2 lines)

The vector b1 × b2 does not have to be perpendicular to the vector from the origin to P. You can construct a system of three equations P + u(b1 × b2) = Q where u is a real number and find s, t and u (I get s = 1, t = 5 and u = -1). Searching the web for "distance between skew lines" also returns many results, such this this PDF.