Distance between two lines

Find an expression for the square of the distance between any two points on

L_1

and

L_2

in terms of

\mu

and

\lambda

where

L_1

has equation

\mathbf{r} =\displaystyle \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} +\lambda \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}

and

L_2

has equation

\mathbf{r} = \displaystyle \begin{pmatrix} 1 \\ -4 \\ 6 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}

.

Without using the scalar product and without using a standard shortest distance formula, find the shortest distance between the lines

L_1

and

L_2

and give the coordinates of each of the points on

L_1

and

L_2

at which the shortest distance occurs.

This question is proving quite difficult. It is an FP4* question.I would normally use scalar and/or vector product to do this but it specifically states that you can't.

(edited 7 years ago)

Original post by Ano123

Find an expression for the square of the distance between any two points on

L_1

and

L_2

in terms of

\mu

and

\lambda

where

L_1

has equation

\mathbf{r} =\displaystyle \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} +\lambda \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}

and

L_2

has equation

\mathbf{r} \displaystyle \begin{pmatrix} 1 \\ -4 \\ 6 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}

.

Without using the scalar product and without using a standard shortest distance formula, find the shortest distance between the lines

L_1

and

L_2

and give the coordinates of each of the points on

L_1

and

L_2

at which the shortest distance occurs.

This question is proving quite difficult. It is an FP4* question.I would normally use scalar and/or vector product to do this but it specifically states that you can't.

Surely you can use the cross/vector product? That wouldn't be the same as using the scalar product and it would give you a nice perpendicular vector to both lines.

(edited 7 years ago)

Reply 2

Ano123

OP

15

Original post by RDKGames

Surely you can use the cross/vector product? That wouldn't be the same as using the scalar product and it would give you a nice perpendicular vector to both lines.

That way is practically a standard formula, or you would have to use the scalar product after that wouldn't you?

Reply 3

RichE

15

Original post by Ano123

Find an expression for the square of the distance between any two points on

L_1

and

L_2

in terms of

\mu

and

\lambda

where

L_1

has equation

\mathbf{r} =\displaystyle \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} +\lambda \begin{pmatrix} -1 \\ 2 \\ 3 \end{pmatrix}

and

L_2

has equation

\mathbf{r} \displaystyle \begin{pmatrix} 1 \\ -4 \\ 6 \end{pmatrix} + \mu \begin{pmatrix} -1 \\ -2 \\ 1 \end{pmatrix}

.

Without using the scalar product and without using a standard shortest distance formula, find the shortest distance between the lines

L_1

and

L_2

and give the coordinates of each of the points on

L_1

and

L_2

at which the shortest distance occurs.

This question is proving quite difficult. It is an FP4* question.I would normally use scalar and/or vector product to do this but it specifically states that you can't.

If you call the first point P and the second point Q, then |PQ|^2 is a function in lambda and mu which is a degree two function of two variables. Has this question come up at all in the context of conics? Or diagonalizing symmetric matrices?

PS Actually it's easier than that - just work out |PQ|^2

(edited 7 years ago)

Original post by Ano123

That way is practically a standard formula, or you would have to use the scalar product after that wouldn't you?

In that case I think differentiation might do the trick. I used it once for distances between lines and planes in a 9 marker and got it right, it wasn't on the mark schemes though. I'll give it a shot and see what happens.

Never mind, that works when you actually know the distance. Useful for finding the points though. Do you have the answers for this? Been a while since I've done FP4 :smile:

(edited 7 years ago)

Reply 5

DylanJ42

18

Original post by Ano123

x

okay for the first part i think you just just use the 3d version of Pythagoras' theorem (im not sure if this has it's own name or not) with your first set of coordinates being

\displaystyle \left( 2 - \lambda,\: -1 + 2\lambda,\: 4 + 3\lambda \right)

and your second being

\displaystyle \left(1 - \mu ,\: -4 - 2\mu,\: 6 + \mu \right)

. Theres some tidying up to do though.

Note: I havent tired this method for the next part yet and im very tired but im guessing itll work

so I have no idea how to differentiate with both lambda and mu in there, but you know the shortest distance between two lines that dont touch is the perpendicular line to both of them so I'd use vector product for this part.

once you get your direction vector which is perpendicular to both lines, lets call it

\displaystyle \begin{pmatrix} a, \ b, \ c \end{pmatrix}

, just use the coordinates for a general point on one line (use the things in the first paragraph) and then;

\displaystyle \left( 2 - \lambda,\: -1 + 2\lambda,\: 4 + 3\lambda \right)

+ t(a,b,c) =

\displaystyle \left(1 - \mu ,\: -4 - 2\mu,\: 6 + \mu \right)

Now youll have three equations with three unknowns (namely lambda, mu and t) so solve for t and that'll lead to finding the shortest distance, solve for lambda and mu to find the exact points on the lines they are referring to

Reply 6

Ano123

OP

15

I did it.
If we let

\displaystyle \mathbf{d}= \mathbf{r}_1 - \mathbf{r}_2 = \begin{pmatrix} 1-\lambda - \mu \\ 3+2\lambda -2\mu \\ 3+3\lambda -\mu \end{pmatrix}

.
Let the distance be

D

so

D^2= |\mathbf{d} |^2 = (1-\lambda -\mu )^2 +(3+2\lambda -2\mu )^2+(3+3\lambda -\mu )^2.

So

D^2= 6\mu ^2 +18 \mu +14\lambda ^2 -2\lambda +14

. We can treat it as a quadratic in

\mu

and then complete the square.

D^2 = 6 \left (\mu +\frac{3}{2} \right )^2 +14 \left (\lambda -\frac{1}{14} \right )^2 +3/7

.
Making both brackets 0 we see that for minimum

D^2

,

\lambda =1/14

and

\mu =-3/2

and so

D_{\text{min}} = \frac{1}{7} \sqrt{21}

.

(edited 7 years ago)

Reply 7

B_9710

18

Alternatively, define

k(\mu , \lambda )=6\mu ^2+18\mu + 14\lambda ^2 - 2\lambda +14

.
We have

\displaystyle \frac{\partial k}{\partial \mu }=12\mu +18

and

\displaystyle \frac{\partial k }{\partial \lambda }=28\lambda -2

.
Setting both derivatives to 0 we obtain

\mu =-3/2

and

\lambda =1/14

for

k

to have a minimum value.

Distance between two lines

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