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Given lines $\mathbb{L}_1 : \lambda(1,3,2)+(-1,3,1)$, $\mathbb{L}_2 : \lambda(-1,2,3)+(0,0,-1)$ and $\mathbb{L}_3 : \lambda(1,1,-2)+(2,0,1)$, find a line $\mathbb{L}$ such that $\mathbb{L}$ is parallel to $\mathbb{L}_1$, $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$ and $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$.

Since $\mathbb{L}$ must be parallel to $\mathbb{L}_1$, then $\mathbb{L}:\lambda(1,3,2)+(x,y,z)$ but I can't figure out how to get that (x,y,z) point. I'd like to be given just a slight nod because I'm sure the problem is really easy. Thanks a lot!

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In English: find the line parallel to $\mathbb{L}_1$ that crosses the other two given lines. –  J. M. Oct 2 '10 at 4:22
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4 Answers 4

up vote 5 down vote accepted

Your first step is correct. Let's write the vector equations of your line $\mathbb{L}$ and the remaining ones:

$$ \begin{align} \mathbb{L}_{\ \ } \qquad (x,y,z) &= (x_0, y_0, z_0) + \lambda (1, 3,2) \\ \mathbb{L}_2 \qquad (x,y,z) &= (0,0,-1) + \lambda_2 (-1, 2,3) \\ \mathbb{L}_3 \qquad (x,y,z) &= (2,0,1) + \lambda_3 (1,1,-2) \end{align} $$

Since $\mathbb{L}\cap\mathbb{L}_2 \neq \emptyset$, you may take the point $(x_0, y_0, z_0) \in \mathbb{L}_2$. So the equation for $\mathbb{L}$ must look like

$$ \mathbb{L} \qquad (x,y,z) = (-\lambda_2 , 2\lambda_2, -1 + 3\lambda_2 ) + \lambda (1, 3,2) \ . $$

As for $\mathbb{L}\cap\mathbb{L}_3 \neq \emptyset$, you should find the common point thanks to

$$ (-\lambda_2 , 2\lambda_2, -1 + 3\lambda_2 ) + \lambda (1, 3,2) = (2,0,1) + \lambda_3 (1,1,-2) $$

Somewhat in disguise, this is a $3\times 3$ system of linear equations, with unkowns $\lambda, \lambda_2$ and $\lambda_3$. Solve it. In particular, find $\lambda_2$ and you have your line $\mathbb{L}$.

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Your line is not parallel to $\mathbb{L}_1$ as you've accidently used (-1,3,1) instead of (1,3,2). –  Derek Jennings Oct 2 '10 at 8:07
    
@fmartin: Just to let you know you have the answer, any of (2-w,-4-3w,-7-2w), for some w, will do for your point (x,y,z). –  Derek Jennings Oct 2 '10 at 8:14
    
@Derek. Thanks. Corrected. –  a.r. Oct 2 '10 at 8:16
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I'd like to add the following to Agustí Roig's splendid answer.

The equation of the general line passing through both $\mathbb{L}_2$ and $\mathbb{L}_3$ is given by $\underline{r} = \underline{a}+ \mu(\mathbb{L}_2 - \mathbb{L}_3),$ where $\underline{a}$ is some point on this line and its direction is $\mathbb{L}_2 - \mathbb{L}_3.$

Now we know that the direction is (1,3,2), so set $\lambda (1,3,2) = \mathbb{L}_2 - \mathbb{L}_3.$ That is $$\lambda (1,3,2) = (0,0,-1)+\lambda_2(-1,2,3) – (2,0,1) - \lambda_3(1,1,-2).$$

Hence we obtain $\lambda_2 = -2$, $\lambda_3 = 2$ and $\lambda = -2.$

Putting $\lambda_2 = -2$ in the equation of $\mathbb{L}_2$ gives the point (2,-4,-7), which is equivalent to w=0 in my comment to Agustí's answer. I hope that this adds some value for you.

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$\mathbb L \cap \mathbb L_2 \ne \emptyset$ implies that for some value of $\lambda$, your equation for $\mathbb L$ is on $\mathbb L_2$. Since substituting $\lambda' = \lambda - c$ doesn't change the line $\mathbb L$, you can take $\lambda = 0$ without loss of generality. That is, $(x,y,z)$ is on $\mathbb L_2$, or $(x,y,z) = \lambda_2(-1,2,3) + (0,0,-1)$. ...

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There is something fishy about you answer, isn't it? -You don't seem to need the third line at all. And indeed your solution for the line $L$ doesn't cross the third one. Your mistake is the following: $L$ crosses $L_2$, right, but who did tell you that it crosses at (0,0,-1)? –  a.r. Oct 2 '10 at 7:51
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This isn't a complete solution. It's just the first step, since the OP asked for "a slight nod". –  aschepler Oct 2 '10 at 13:28
    
Sorry, I misunderstood what you were saying. –  a.r. Oct 2 '10 at 14:27
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Notation can help us see what's going on. Let's systematically parametrize the lines: let $\mathbb{L_2}$ start at $\mathbf{a} = (0,0,-1)$ and have a direction vector $\mathbf{u} = (-1,2,3)$ and similarly let $\mathbf{b} = (2,0,1)$, $\mathbf{v} = (1,1,-2)$ determine $\mathbb{L_3}$. All we need to know about $\mathbb{L_1}$ is its direction vector, $\mathbf{w} = (1,3,2)$.

If there exists a solution then we can start at the origin $\mathbf{a}$ of $\mathbb{L_2}$ and reach the origin $\mathbf{b}$ of $\mathbb{L_3}$ via a path that takes us along $\mathbb{L_2}$, thence in the direction $\mathbf{w}$ of $\mathbb{L_1}$ to a point on $\mathbb{L_3}$, and finally back along $\mathbb{L_3}$ to $\mathbf{b}$. Let $\lambda$, $\rho$, and $\mu$ be the multiples of the direction vectors $\mathbf{u}$, $\mathbf{w}$, and $\mathbf{v}$ that are traveled, respectively. Whence

$$\mathbf{b - a} = \lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}$$

and the solution line $\mathbb{L}$ passes through $\mathbf{a} + \lambda \mathbf{u}$ in the $\mathbf{w}$ direction. All we need to do, then, is find $\lambda$. A simple way is to find a linear form that kills both $\mathbf{w}$ and $\mathbf{v}$ but not $\mathbf{u}$ and apply it to both sides. The obvious one is to take a dot product with some vector that is mutually perpendicular to $\mathbf{w}$ and $\mathbf{v}$. All vectors are multiples of the cross product $\mathbf{v} \times \mathbf{w}$, so let's use it. Thus

$$(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w}) =(\lambda \mathbf{u} + \rho \mathbf{w} + \mu \mathbf{v}) \cdot (\mathbf{v} \times \mathbf{w}) = \lambda \mathbf{u} \cdot (\mathbf{v} \times \mathbf{w}).$$

Dividing gives the solution (if it exists!)

$$\lambda = \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} = -2.$$

A parametrization of the line $\mathbb{L}$ therefore is

$$\mathbf{a} + \mathbf{u} \frac{(\mathbf{b - a}) \cdot (\mathbf{v} \times \mathbf{w})}{\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})} + t \mathbf{w}$$

for a real parameter $t$. To check the solution, reverse the roles of the lines $\mathbb{L_2}$ and $\mathbb{L_3}$, thereby finding $\mu$ and the point along $\mathbb{L_3}$. (We don't have to recompute the triple product in the denominator; only its sign changes.) The corresponding point (on $\mathbb{L_3}$ now) indeed lies on the solution line (with $t = 2$).

This approach immediately shows a solution does exist and is unique if and only if the triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is nonzero; i.e., the three direction vectors $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are not coplanar.

Of course this all is just another way of reproducing the previous solutions. Its possible advantages are its geometric clarity and the straightforward formulas it provides for the answer.

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Wow, the box product makes everything tidy! :D –  J. M. Oct 3 '10 at 0:49
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