Perpendicular Distance Between the Lines

Find the (perpendicular) distance between te lines

L1 : x= -1 + t L2 : x = -1
y = -1 + t y = 1
z = 1 + t z = t

First I decided that the 2 lines dont intersect

since setting

x= -1 + t = -1
y = -1 + t = 1
z = 1 + t = t'

there is no value of t which satisfies

A normal vector to both L1 and L2 is (1,-1,0),

{which i calculated using the Cross Produect ( ie that of the parallel vectors of each respective line (1,1,1) and (0,0,1) )}

Using a point on L1 as (-1,-1,1) and denote it "A" and choosing a point on L2 (-1,1,0) denoting it "P", I have AP = (0,2,-1).

Hence using |proj AP onto n| = | AP.n^| = |AP.(n/|n|)| = Perpendicular Distance

Therefore Equalling = (0,2,-1) . [(1,-1,0)/sqrt(2)]

= |-2 / [sqrt(2)]| or simply the sqrt(2)


IS MY ANSWER CORRECT??? (sorry it took me so long, just wanted to include all my working so if a make a mistake somewhere could you show me where my undersatnding lacks!) thanks

The problem is, ans why I think im incorrect, is all questions I have done like this previosuly, the lines intersect, and thus this intersetcion point is used.....

If two lines intersect - the perpendicular distance between those two lines must be equal to zero.

Your method and answer looks good.

umm...no im first year, first semester college....so do the lines intersect???...i thought they didnt??...

No - the lines do not intersect.

If they did intersect the answer (distance) would equal to zero.

cool...as i thought...then do you think my answer is correct??...thankyou for the link to the webpage

My goodness it is next to impossible to read your formatting!
\(\displaystyle L_1 \left\{ {\begin{array}{l}
{x = - 1 + t} \\
{y = - 1 + t} \\
{z = 1 + t} \\
\end{array}} \right.\quad \& \quad L_2 \left\{ {\begin{array}{l}
{x = - 1} \\
{y = 1} \\
{z = t} \\
\end{array}} \right.\)

The common normal is indeed \(\displaystyle N = \left[ {\begin{array}{r}
1 \\
{ - 1} \\
0 \\
\end{array}} \right]\).
Now if we know the common normal, N, and one points point on each line
\(\displaystyle P \in L_1 \quad \& \quad Q \in L_2\) then the distance between the two lines is given by:
\(\displaystyle D\left( {L_1 ,L_2 } \right) = \frac{{\left| {\vec{PQ} \cdot N} \right|}}{{\left\| N \right\|}}\)

Hello, ku1005!

I used Calculus ... just to verify your answer.

Find the (perpendicular) distance between the lines

\(\displaystyle L_1:\;\begin{Bmatrix}x& =& -1\,+\, t \\ y & = & -1\,+\,t \\ z & = & 1\,+\,t\end{array}\;\;\;L_2:\;\begin{Bmatrix}x & = & -1 \\ y & = &1 \\ z & = & u\end{array}\)

Click to expand...

Let \(\displaystyle d\)= distance between the two lines.
Let \(\displaystyle D\) = the square of this distance.

We have: \(\displaystyle \ \;=\;[(-1\,+\,t)\,-\,(-1)]^2\,+\,[(-1\,+\,t)\,-\,1]^2\,+\,[(1\,+\,t) \,-\,u]^2\)

. . . . . . . . \(\displaystyle D \;=\;t^2\,+\,(t\,-\,2)^2\,+\,(t\,-\,u\,+\,1)^2\)

and we will minimize \(\displaystyle D.\)


Set the two partial derivatives equal to zero and solve.

. . \(\displaystyle \frac{\partial D}{\partial t} \;=\;2t\,+\,2(t\,-\,2)\,+\,(2(t\,-\,u\,+\,1) \;=\;0\)
. . \(\displaystyle \frac{\partial D}{\partial u} \;=\;-2(t\,-\,u\,+\,1)\;=\;0\)

We have: \(\displaystyle \:\begin{array}{ccc}3t\,-\,u & = & 1 \\ t\,-\,u & = & -1\end{array}\)

Subtract: \(\displaystyle \:2t\,=\,2\;\;\Rightarrow\;\;t\,=\,1\;\;\Rightarrow\;\;u\,=\,2\)

For \(\displaystyle t\,=\,1\), the point on \(\displaystyle L_1\) is: \(\displaystyle \;P(0,\,0,\,2)\)

For \(\displaystyle u\,=\,2\), the point on \(\displaystyle L_2\) is: \(\displaystyle \:Q(-1,\,1,\,2)\)


Then: \(\displaystyle \;PQ \;=\;\sqrt{(-1-0)^2\,+\,(1-0)^2\,+\,(2-2)^2} \;=\;\sqrt{1\,+\,1\,+\,0} \;=\;\L\sqrt{2}\)

Using vector methods:
\(\displaystyle \L \begin{array}{l}
P = \left\langle { - 1, - 1,1} \right\rangle ,\quad Q = \left\langle { - 1,1,0} \right\rangle \quad \& \quad N = \left\langle {1, - 1,0} \right\rangle \\
\vec {PQ} = \left\langle {0,2, - 1} \right\rangle ,\quad \frac{{\left| {\vec {PQ} \cdot N} \right|}}{{\left\| N \right\|}} = \frac{2}{{\sqrt 2 }} = \sqrt 2 \\
\end{array}.\)

thanks heaps...and sorry about formatting, thats not how i ment for it to show up in the post