Ok so this is basically for my upcoming test. The essence of question is that I have a plane $$E_1: \; 18x-y-10z=1$$ and a line $$g(x)=(1,-3,-2)+x(1,-2,0)$$

What I need to do next is find a plane $E_2$ which is perpendicular to $E_1$ and containing the line described by $g(x)$

Since this is a part of a bigger problem the plane $E_1$ already contains the line $g(x)$

I have tried to do following. Since the line is given I can find any two points which it contains. So I choose $x=0$ and $x=1$ and plug them into function to obtain two points contained by the line. These points are:

$$P_0(1,-3,-2)$$ $$P_1(2,-5,-2)$$

Now since I have two points,I can find a vector between them $$P_0P_1 (1,-2,0)$$

Also I can determine the normal vector of plane $E_1$ which will be denoted $n_1$ and is equal to $$n_1=(18,-1,-10)$$

Now let $n_2$ be the perpendicular vector of plane $E_2$.Since these two planes are perpendicular to each other,then their normal vectors must be perpendicular to each other, and dot product of the normal vectors of two planes must be zero. When I take scalar product I get the first equation on my way:


Next equation I obtain is by taking in consideration that $n_2$ is normal to plane $E_2$ thus normal to contained vector $P_0P_1$ which gives the second equation:


From here on I am so to say, stuck. I cant obtain a second equation, nor can I find a third non-collinear point to finish the system. How do I determine the plane?

  • $\begingroup$ Just found out,that if I solve the equation system with parameters and plug in random x,it does not seem to be the normal vector to plane $E_2$ $\endgroup$ Commented Dec 20, 2015 at 19:35
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    $\begingroup$ You say that $E_1$ contains $g(x)$, but $(1,-3,-2) + (1,-2,0) = (2,-5,-2) \notin E_1$. $\endgroup$ Commented Dec 20, 2015 at 19:40
  • $\begingroup$ I made a mistake in previous part of exercise.Correction coming. $\endgroup$ Commented Dec 20, 2015 at 20:00
  • $\begingroup$ Corrected the plane equation,and second equation which comes out of scalar product. $\endgroup$ Commented Dec 20, 2015 at 20:38
  • $\begingroup$ The line $g$ still isn’t in the plane $E_1$: if you substitute $(t+1, -2t-3, -2)$ into the equation of the plane, you get $20t+41=1$, which isn’t true for all values of $t$. $\endgroup$
    – amd
    Commented Dec 20, 2015 at 21:23

2 Answers 2


You have two vectors parallel to $E_2$: $v=(1,-2,0)$ and $n_1=(18,-1,-10)$, so you can write immediately the parametric equations for $E_2$: $$ (x,y,z)=P_0+sv+tn_1=(1+s+18t,-3-2s-t,2-10t). $$ The cartesian equation can be readily found from these: express $t$ and $s$ using the equations for $z$ and $y$, then substitute into the equation for $x$.

  • $\begingroup$ This pretty concise,I did not know that we can do that.Do you happen to have acces to further resources where I can find a proof ,or derivation of this fact? $\endgroup$ Commented Dec 21, 2015 at 16:40
  • $\begingroup$ Do you mean the parametric equation of a plane? See here for instance: math.stackexchange.com/questions/152467/… $\endgroup$ Commented Dec 21, 2015 at 16:58
  • $\begingroup$ Let me add that you can avoid the first step in your solution (finding vector $P_0P_1$) because a vector directed along the line is already there in your parametric equation, multiplied by the parameter: $(1,-2,0)$. In addition, line $g$ does not lie on the plane. $\endgroup$ Commented Dec 21, 2015 at 17:03
  • $\begingroup$ No I meant that if I have two vectors parallel to a plane,that I can just plug them into the parametric equation. $\endgroup$ Commented Dec 21, 2015 at 17:52
  • $\begingroup$ Well, that's quite obvious, watch this video for instance. Of course the two vectors must be linearly independent (i.e. they must not be parallel). $\endgroup$ Commented Dec 21, 2015 at 18:34

You can make the computations simpler by moving everything to the origin, solving, and then translating back.

Assuming that the line $\mathbf g(t)=\mathbf g_0+\mathbf wt$ is indeed in the plane $E_1$, start by translating by $-\mathbf g_0$. This should give you an equation of the plane of the form $\mathbf n\cdot\mathbf x=0$. The normal to the plane you seek will be perpendicular to both the line $\mathbf g$ and the normal $\mathbf n$ to the plane $E_1$, so the equation $(\mathbf n\times\mathbf w)\cdot\mathbf x=0$ gives you that plane. Translating it by $\mathbf g_0$ should give you the solution to the original problem.


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