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This is how the question was given to me:

Find the equation of a plane that passes through the line $$p: \frac {x-3}{2} = \frac {y+4}{1}= \frac{z-2}{-3}$$ and is parallel to another line $$ q: \frac {x+5}{4} = \frac {y-2}{7}= \frac{z-1}{2} $$

So I know if a plane is parallel to a line that means the dot product of the normal vector of that plane and the vector given by the line $q=(4,7,2)$ should be zero. So that gives me one equation: $4x+7y+2z=0$.

And if the line p passes through the plane then by substituting the equations given for $x$, $y$, $z$ (from the parametric equations: $p:x=2t+3; y=t-4; z=-3+2$) I can get another equation: $A(2t+3)+B(t-4)+C(-3t+2)$

If I combine the two I get that $t=\frac {12}{9}$ which doesn't seem right. Also I have no idea what to do further than that. Where did I go wrong?

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4 Answers 4

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Probably, the best approach is that you find first a point $P$ on $p$ ( simply put e.g. $z=2$ and get $y$ and $x$ so $P=(3,-4,2)$).

Then construct the equation of a plane passing through $P$ with normal vector equal to the cross-product of the direction vectors of $p$ and $q$ :$(23,-16,10)$.

That is

$$ 23\left( {x - 5} \right) - 16\left( {y + 3} \right) + 10\left( {z + 1} \right) = 0\quad \Rightarrow \quad 23x - 16y + 10z - 153 = 0 $$

and if you substitute in it the parametric equations for $p$

$$ x = 2t + 3,\quad {\rm }y = t - 4,\quad {\rm }z = - 3t + 2 $$ you get $0=0$, and of course you have $$ \eqalign{ & 23 \cdot 2 - 16 \cdot 1 + 10 \cdot \left( { - 3} \right) = 0 \cr & 23 \cdot 4 - 16 \cdot 7 + 10 \cdot 2 = 0 \cr} $$

Going your way, a general plane parallel to $q$ is $q_x(x-x_0)+ \cdots =0 $, i.e. $q_x x+ \cdots=d$.
If in it you put the parametric equations (in $t$) of $p$, since $p$ lies on the plane, the plane equation shall be satisfied for whichever value of $t$, which means that you shall find the $d$ which leads you to $0t=0$.

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  • $\begingroup$ Okay, so then $n=p x q= (23,-16,10)$ and an intersection point $P$ is $(1, -3, -1)$. So I get the equation for the plane as $23x-16y+10z-56=0$ once I solve for d. Is that what you meant? Because looking at them on GeoGebra the plane and q don't look too parallel to me. $\endgroup$ Mar 10, 2018 at 21:02
  • $\begingroup$ what is the point $(1, -3, -1)$ ? it doesn't lie on $p$. $\endgroup$
    – G Cab
    Mar 10, 2018 at 21:15
  • $\begingroup$ Oh, I suppose it doesn't. Shouldn't any point that solves the equation lie on the line? $\endgroup$ Mar 10, 2018 at 21:25
  • $\begingroup$ any point yes , BUT on $p$ (since it is $p$ the line that lies on the plane) . I've added an example. $\endgroup$
    – G Cab
    Mar 10, 2018 at 21:28
  • $\begingroup$ Yes, I meant any point the solves the equation for p. $\endgroup$ Mar 10, 2018 at 21:29
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use that the plane has the equation $$ax+by+cz+d=0$$ and since the line should be parallel to the plane the dot-product of $$\vec{n}=[a;b;c]$$ and the direction vector of the second line $$\vec{a}=[4;7;2]$$ must be zero and we get $$4a+7b+2c=0$$ since the plane containes the first line we get $$a(2t+3)+b(t+4)+c(-3t+2)+d=0$$ simplifying this equation $$t(2a+b-3c)+3a+4b+2c+d=0$$ then we get $$2a+b-3c=0$$ and $$3a+4b+2c+d=0$$ can you finish?

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  • $\begingroup$ Yes, but I don't know how to use that information to get the equation for the plane $\endgroup$ Mar 10, 2018 at 20:52
  • $\begingroup$ i have added some lines for you! $\endgroup$ Mar 10, 2018 at 20:57
  • $\begingroup$ I'm sorry, I don't quite follow. How would that five me an equation for a plane? Isn't that how I'd find whether two planes intersect? $\endgroup$ Mar 10, 2018 at 21:06
  • $\begingroup$ no you must compute the variables $a,b,c,d$ these are the coefficients of the plane $\endgroup$ Mar 10, 2018 at 21:09
  • $\begingroup$ Ah, I think I understand. Thank you $\endgroup$ Mar 10, 2018 at 21:14
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Note that

  • line $p$ has equation $(3,-4,2)+t(2,1,-3)$
  • line $q$ has equation $(-5,2,1)+t(4,7,2)$

Thus a plane containing $p$ has equation $ax+by+cz+d=0$ such that $2a+b-3c=0$ that is

  • $ax+(3c-2a)y+cz+d=0$

From the condition that it contains the point $(3,-4,2)$we obtain

  • $3a-12c+8a+2c+d=0\implies d=10c-11a$

  • $ax+(3c-2a)y+cz+10c-11a=0\implies x+(3k-2)y+kz+10k-11=0$

to be parallel to plane $q$ normal vector must be orthogonal to direction vector of $q$ that is

  • $4+21k-14+2k=0\implies k=\frac{10}{23}$

therefore

$$x-\frac{16}{23}y+\frac{10}{23}z+\frac{100}{23}-11=0\iff23x-16y+10z-153=0$$

In short the plane equation $ax+by+cz+d=0$ can be obtained assuming $(a,b,c)=v_p\times v_q$ and finding $d$ by the condition that $(3,-4,2)$ belongs to it.

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  • $\begingroup$ What exactly happens in this step $ax+(3c−2a)y+cz+d=0$? $\endgroup$ Mar 10, 2018 at 21:45
  • $\begingroup$ We have plugged $b=(3c-2a)$. $\endgroup$
    – user
    Mar 10, 2018 at 21:49
  • $\begingroup$ Oh, I see! This is an interesting method. Thank you! $\endgroup$ Mar 10, 2018 at 21:53
  • $\begingroup$ @crylikeacanary It is useful to touch what it is going on step by step! $\endgroup$
    – user
    Mar 10, 2018 at 21:54
  • $\begingroup$ You substituted '1' for 'a' in the line 'ax+(3c-2a)y+cz+10c-11a'. Could you pls explain this, why '1'? $\endgroup$
    – twa14
    Mar 1, 2021 at 11:37
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This is pretty straightforward if you take a projective-geometric approach and use homogeneous coordinates. A parameterization of the first line is $(3,-4,2)+\lambda(2,1,-3)$. The homogeneous coordinates of two points on the line can be read directly from this parameterization (and also from the original equations) $[3:-4:2:1]$ and $[2:1:-3:0]$. The latter is a point at infinity that corresponds to the line’s direction vector. The plane must be parallel to $q$, which means that their intersection is the point at infinity that corresponds to $q$’s direction vector, so we have a third point on the plane, $[4:7:2:0]$.

For any point $\mathbf x$ on the plane $\mathbf\pi$ we must have $\mathbf\pi\cdot\mathbf x=0$. Applying this to the three known points generates a system of three linear equations in the components of $\mathbf\pi$, from which we see that $\mathbf\pi$ is a null vector of the matrix $$\begin{bmatrix}3&-4&2&1\\2&1&-3&0\\4&7&2&0\end{bmatrix}.$$ A straightforward row-reduction gives $[23:-16:10:-153]$, which corresponds to the equation $$23x-16y+10z=153.$$

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  • $\begingroup$ This is beyond anything we've covered in class, but it is the same answer I got! $\endgroup$ Mar 10, 2018 at 21:44
  • $\begingroup$ @crylikeacanary I like this method because it’s quite direct and only requires knowing how to find the null space of a matrix, which is a basic operation in linear algebra. It also extends directly to higher-dimensional spaces. $\endgroup$
    – amd
    Mar 10, 2018 at 21:48
  • $\begingroup$ Yes, I can see what the advantage of solving it this way is. $\endgroup$ Mar 10, 2018 at 21:55

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