How to obtain the equation of a plane which intersects with a given plane in a given line

Question

Given the plane $2x + y - 3z = 1$. How to obtain the equation of a plane which intersects this plane in the line $$r=\left( \begin{array}{c}1\\2\\1\\\end{array}\right)+t\left( \begin{array}{c}1\\-2\\0\\\end{array}\right)?$$

I have assumed that the other plane is $ax+by+cz=d$. Then the line satisfies this plane $$a(1+t)+b(2-2t)+c=d.$$ But I can't proceed from here.

Answer 1

I will follow your idea. From $$a(1+t)+b(2-2t)+c=d$$ you get

$$at-2bt+a+2b+c=d.$$ To have the equality it is obvious that $a=2b.$ So you have the relation

$$4b+c=d.$$

Thus, any plane of the form

$$2bx+by+cz=4b+c$$ solves the problem. (Well, you have to exclude $2x+y-3z=1$.)

Answer 2

For example, take $$(x,y,z)=(1,2,1)+t(1,-2,0)+s(1,0,1),$$ which gives $$2x+y-2z-2=0.$$

Answer 3

For the new plane we need that the normal vector $\vec n=(a,b,c)$ is orthogonal to the given line that is

$$(a,b,c)\cdot (1,-2,0)=0 \implies (a,b,c)=(2,1,0)$$

and we obtain

$$2x+y+d=0$$

then we can find $d$ by the condition that $P=(1,2,1)$ belongs to the plane, that is

$$2+2+d=0\implies d=-4$$

therefore

$$2x+y-4=0$$

is a possible solution.

Answer 4

If all you know is that the line $r$ lies in the desired plane, then the plane is not unique. If we find one such plane, we can rotate it about the line $r$ to obtain another plane in which $r$ lies. This means that there is a one-parameter family of planes that could solve the problem.

The fact that the line $r$ lies in another plane is immaterial to the question, unless for some reason $r$ doesn't actually lie in the other plane (in which case there is no plane which intersects the other plane in $r$.)

Answer 5

A plane $P$ is an answer two your question if and only if:

1. $P$ isn't the plane $2x+y-3y=1$;
2. $r\subset P$.

Now, a plane $P$ defined by $ax+by+cz=d$ contains $r$ if and only if $(1,2,1),(2,0,1)\in P$. This means that$$\left\{\begin{array}{l}a+2b+c=d\\2a+c=d,\end{array}\right.$$which is equivalent to $a=2b$ and $c=-4b+d$. So, the answer is: the planes of the form$$2bx+by+(-4b+d)z=d,$$for which there's no $\lambda\in\mathbb R$ such that $\lambda(2,1,-3,1)=(2b,b,-4b+d,d)$.

Answer 6

Try to think geometrically rather than algebraically. You're given a plane $\Pi$; think about this plane as determined by a reference point and a normal vector $\mathbf{n}$. You're also given a line $\ell$ within that plane; that's determined by a reference point and a direction vector $\mathbf{v}$. Since $\ell$ is contained in $\Pi$, we can choose the reference points of each to be the same point $P$, and we know that $\mathbf{n}$ and $\mathbf{v}$ are orthogonal.

We want another plane $\Pi'$; again, it's determined by a reference point and normal vector $\mathbf{n}'$. It's required to include $\ell$, so it must contain $P$, and $\mathbf{n}'$ must be orthogonal to $\mathbf{v}$ too. We might as well use $P$ as the reference point for $\Pi'$. As long as $\mathbf{n}'$ is not parallel to $\mathbf{n}$ (otherwise the planes coincide), their intersection will be $\ell$ alone.

It remains to find a vector $\mathbf{n}'$ that is perpendicular to $\mathbf{v}$ and not parallel to $\mathbf{n}$. Now it's an algebra problem, but let's keep it in terms of vector algebra. There are infinitely many choices, but one convenient one is $\mathbf{n}' = \mathbf{n} \times \mathbf{v}$. This is nonzero because $\mathbf{n}$ and $\mathbf{v}$ are nonzero orthogonal vectors, and it's orthogonal to both $\mathbf{n}$ and $\mathbf{v}$. In particular, it's not parallel to $\mathbf{n}$.

Therefore a solution to your problem is the plane through $P$ normal to $\mathbf{n} \times \mathbf{v}$, where $P$ is a point on $\ell$, $\mathbf{v}$ is a direction vector of $\ell$, and $\mathbf{n}$ is a normal vector to $\Pi$. And we did this without any messy algebra!

But you probably need an equation for this plane. Remember $P$ can be any point on $\ell$. Since $\ell$ is parametrized by a vector function $\mathbf{r}(t)$, a convenient point to call $P$ is the tip of $\mathbf{r}(0)$. That is, let $P = (1,2,1)$. Since $\Pi$ has equation $2x + y - 3z = 1$, the vector of coefficients $\mathbf{n} = \left<2,1,-3\right>$ is a normal vector to $\Pi$. Finally, a direction vector $\mathbf{v}$ to $\ell$ can be found by taking $\mathbf{r}'(t) = \left<1,-2,0\right>$. Therefore, $$ \mathbf{n}' = \mathbf{n} \times \mathbf{v} = \left<-6,-3,-5\right> $$ And $\Pi'$ has equation \begin{align*} -6(x-1) - 3(y-2) - 5(z-1) &= 0 \\\implies -6x -3y - 5z &= -17 \\\implies 6x + 3y + 5z &= 17 \end{align*}

Answer 7

There are infinitely many such planes e.g. one is the plane itself and the other is the perpendicular plane. If we denote the characteristic vector of the unknown plane by $(a,b,c)$ it's obvious that it should be normal on the characteristic vector of the given line, (1,-2,0) which means that $$a-2b=0$$so the vector is $$(2b,b,c)$$ but the plane also includes (1,2,1) which means that $$2bx+by+cz=4b+c$$if $b\ne0$ we have $$2x+y+\lambda z=4+\lambda\quad,\quad \lambda\in\Bbb R$$which is one set of planes. The other plane is $z=1$