# Find the equation of the intersection of two tangent planes

I have two tangent planes that I've worked out from a previous part of this question:

$$(1): 2x-2y+z=-2$$

$$(2): -8x+2y+4z=18$$

I want to find the parametric equation for the intersection of these two tangent planes at the point $$(-1,1,2)$$ (I don't think the point is necessary anymore, I used this point to find the tangent planes previously).

Here's what I've done:

First I decided to re-write both equations with $$y$$ as the subject so I had:

$$(1): y = x+\frac{1}2 z+1$$

$$(2): y = 9+4x-2z$$

Since I had both equations equal to $$y$$ I made them equal to each other:$$(3):$$ $$9+4x-2z=x+\frac{1}2 z+1$$

From here I tried to find $$z$$ and $$y$$ in terms of $$x$$ so that I could set $$x$$ to be the parameter $$t$$, and then I would have $$z$$ and $$y$$ equal to some form of this parameter $$t$$ as well, however, along the way I think I've made a mistake or two that have led me to the incorrect answer, here's what I did:

First I tried to make $$z$$ the subject in equation $$(3)$$: $$8+3x-\frac{5}2z = 0$$ => $$z=\frac{16}5 + \frac{6}5 x$$ (This doesn't seem quite right to me, other examples I've done very rarely have terms equal to fractions)

Then I took equation $$(1)$$ and since I had $$z$$, I substituted it in: $$2x-2y+\frac{16}5 +\frac{6}5 x = -2$$ => $$y=\frac{16}{10} x +\frac{26}{10}$$

From here I would set $$x =$$ the parameter $$t$$ and then re-write my equations of $$z$$ and $$y$$ to be in terms of $$t$$ and not $$x$$ then I would have my answer but the parametric equation is not correct.

I have access to the answer sheet and the answer provided is:$$(-1,1,2) + (-10,-16,-12)t$$ which is far from what I would get from my equations for $$x,y,z$$.

If someone could show me where I went wrong or if there is a better method to use it would really help.

• Your work appears correct; when $x=-1$, the point is $(-1,1,2)$ Aug 11, 2021 at 13:12
• Then I guess I'm just not sure how to get my answer in that form, my parametric equations would look something like: $x=t$, $y=\frac{16}{10} t + \frac{26}{10}$ and $z=\frac{16}{5} +\frac{6}{5} t$, I'm really not sure these are correct. Aug 11, 2021 at 13:17
• Wait, I'm so blind. If I just multiply all my $t$ values by $10$ I get the correct answer... Aug 11, 2021 at 13:18
• But am I allowed to just multiply by $10$? Also, do I just ignore the non-$t$ terms in the equations for $z,y$? Aug 11, 2021 at 13:19
• There's more that one way to parameterize a line. You could write $P+tV$ where $P$ is a point on the line and $V$ a direction vector, or you could write $tP+(1-t)Q$ where $P$ and $Q$ are two points on the line. In either of those forms, there are many choices, of course. Aug 11, 2021 at 13:20

As already pointed out in the comments, there is no unique parametric representation of a line. In fact, there's infinitely many because you can take any point $$P$$ on the line and any direction vector $$V$$ (any non-zero multiple of any given direction vector, is still a direction vector of the line) to form $$P+tV$$.

You can look at your question as a system of two (linear) equations in three unknowns and solve it with any method you like. The solution, written parametrically, will take the form of the parametric equation of a line - the line of intersection.

(...) or if there is a better method to use it would really help.

One fast way to derive a direction vector of that line, since you already had a point $$P$$, is by using the cross (or vector) product. You may know that you can simply read normal vectors of the given planes since $$(a,b,c)$$ is a normal vector of a plane given by $$ax+by+c+d=0$$, so here:

$$(1): 2x-2y+z=-2$$

$$(2): -8x+2y+4z=18$$

normal vectors of the two planes are $$(2,-2,1)$$ and $$(-8,2,4)$$ respectively. The (line of) intersection of the two planes has to be perpendicular to both planes, so its direction (vector) has to be perpendicular to both normal vectors. This is precisely what the cross product produces and you can verify that: $$(2,-2,1) \times (-8,2,4) = (-10,-16,-12)$$ It can be good practice to simplify the direction vector, definitely if there are follow-up questions which would require further calculations with this result (e.g. divide by $$-2$$).

• Thanks a lot, the method you've shown on how to quickly get the derivative is the method I think I remember my lecturer using to answer this question previously but it was a long time ago so I forgot it. Thanks for the quick reply and the clarification Aug 11, 2021 at 13:50
• This cross product method only applies in 3 dimensions. More generally, your approach is used though using matrix row reduction for efficiency.
– Paul
Aug 11, 2021 at 13:54
• I have just one more question, in your answer you showed a way to quickly find the intersection since it is perpendicular (by using the cross product), In a later part of this question I have to find a point where the $(1)$ is parallel to a given plane (different to the one in this question). I'm just wondering if there's a quick way to find the point where the two are parallel or if a longer method is required. If you want I can create a new post and link it here so you can see what I mean. Aug 11, 2021 at 14:07
• It's not completely clear to me what the (new) question is. With "(1)", are you referring to the first plane? What about the line of intersection? Perhaps it's better to post a new question or if it's something small, add it to this question. Aug 11, 2021 at 14:25
• I'll make a new question and link it here, give me a minute Aug 11, 2021 at 14:35