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Write the equations of the line that passes through point $M(1,0,7)$, is parallel with the plane $3x-y+2z-15=0$ and intersects line $\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}$

Alright, so from what I know, if a line is parallel to a plane, that means that the plane's normal vector is perpendicular to the line's direction vector, so their dot product must be 0. The plane's normal vector is $v=\langle3,-1,2\rangle$ from what I can tell. What other conditions are there for a line to intersect another line, in 3D? Also, can I get an equation from the given point as well?

EDIT:

To get the point of intersection between my given line and the plane, I write down the parametric equations for my line. I have $z=t, y=2t+3, x=4t+1$. I substitute these in the plane equation that you've given me(in which I have my line) and I get the point of intersection between the plane and the line $t=\frac{17}{13}$. That point should be on the line I'm looking for as well. I get point $B(\frac{68}{13},\frac{37}{13},-\frac{74}{13})$.

Then, when I try to compute the line equation from the two points I know are on it, I get $x=\frac{81}{13}+\frac{68}{13}t,\ y=\frac{37}{13}+\frac{37}{13}t,\ z=\frac{17}{13}-\frac{74}{13}t$. The thing is, when I try to compute the dot product between the line's direction vector and the normal vector of the plane to which it is parallel to, I don't get $0$. I ran this many times, and I get the same result every time. Am I doing something wrong? I don't think it's a computation error.

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Another condition "for a line to intersect another line, in 3D" is that there is a point that is on both lines. That sounds simplistic but that really is the condition.

Here is one way to solve your problem. The line that you want is in a plane that is parallel to the plane $3x-y+2z-15=0$. Any such plane has the equation

$$3x-y+2z+D=0$$

for some constant $D$. That plane must also go through the point $M(1,0,7)$, so we can substitute to find the value of $D$:

$$3\cdot 1-0+2\cdot 7+D=0$$

So $D=-17$ and the plane is $3x-y+3z-17=0$.

Combining that with your equations $\frac{x-1}{4}=\frac{y-3}{2}=\frac{z}{1}$ you can find the single point that is the intersection of that plane and your given line. The line you want goes through those two points.

You should be able to finish from here.


There are multiple ways to find the intersection of the plane and of the given line. You showed one way, by giving a parameterization of the line and solving for the parameter. You did make a computation error: the equation you get is

$$3(4t+1)-(2t+3)+2(t)-17=0$$

and the solution is

$$t=\frac{17}{12}$$

which makes $x=\frac{20}3,\ y=\frac{35}6,\ z=\frac{17}{12}$. Substituting this solution into all the equations checks, so this is the right answer. Thus the intersection point is

$$Q\left(\frac{20}3,\ \frac{35}6,\ \frac{17}{12}\right)$$

I got the intersection point by different means, by setting up and solving these simultaneous linear equations:

$$\begin{align} \color{white}{1}x \color{white}{+0y}-4z&=1 \\ y-2z&=3 \\ 3x-y+2z&=17 \end{align}$$

I got the same intersection point.

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  • $\begingroup$ Thanks! I'm still having a bit of an issue with the final result I get. I edited my initial post to outline the steps I've taken. $\endgroup$ – MikhaelM Jan 26 '16 at 18:38
  • $\begingroup$ @MikhaelM: See the addition to my answer. Let me know if you can't finish from there. $\endgroup$ – Rory Daulton Jan 26 '16 at 18:54
  • $\begingroup$ Thank you! Seems I can't add >.< $\endgroup$ – MikhaelM Jan 26 '16 at 19:41

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