linear algebraplane-geometry
I have a question that I could not find on any websites. The question:
Find an equation for the plane that contains the line $x = 1 + t, y = 3t, z = 2t$ and is parallel to the line of intersection of the planes $-x + 2y + z = 0$ and $x + z + 1 = 0$.
I don't know where to start. Can someone give a hint to start me off?
Thanks.
The cleanest way to do this uses the vector product: if $\mathbf{n_1}$ and $\mathbf{n_2}$ are the normals to the planes, then the line of intersection is parallel to $\mathbf{n_1} \times \mathbf{n_2}$.
The equation of a plane takes the form $\mathbf{x.n} = a$, where $\mathbf{x} = (x,y,z)$ and $a$ is a scalar constant. So in your case we have:
$\mathbf{n_1} = (2,-1,1)$ (from $2x−y+z=1$)
$\mathbf{n_2} = (3,1,1)$ (from $3x+y+z=2$)
So the line of intersection is parallel to $\mathbf{n_1} \times \mathbf{n_2} = (-2,1,5)$.
If a plane is given in the so called point-normal form:
$$ax+by+cz=d$$ then the vector
$$\vec n=\begin{bmatrix} a\\ b\\ c \end{bmatrix}$$
is perpendicular to that plane.
We have two planes and the normal vectors are
$$\vec n_1=\begin{bmatrix} 1\\ 1\\ 3\end{bmatrix}\text{ and } \ \vec n_2=\begin{bmatrix} 1\\ -1\\ 2 \end{bmatrix}.$$
Taking the vector product of these two vectors we get a new vector which is perpendicular to both of the normal vectors and it is parallel to both of the planes; and it is then parallel to their intersection line. So,
$$\vec d=\vec n_1\times \vec n_2=\begin{vmatrix} \vec i&\vec j&\vec k\\ 1&1&3\\ 1&-1&2 \end{vmatrix}=\begin{bmatrix} 5\\ 1\\ -2\end{bmatrix}$$
We have the direction vector of the intersection line. Any vector perpendicular to $\vec d$ could serve as a normal vector of a plane that contains the intersection line of the two planes. $\vec n_1$ and $\vec n_2$ would be such vectors but we don't want to use them (or any scalar multiples of them) because we want a plane different from the planes given. In order to find a vector perpendicular to $\vec d$ we need to solve the following scalar product equation
$$\vec d\cdot \vec n_3=\vec d\cdot\begin{bmatrix} x_{n_3}\\ y_{n_3}\\ z_{n_3} \end{bmatrix}=5x_{n_3}+y_{n_3}-2z_{n_3}=0.$$
We may freely choose, say, $x_{n_3}=0$ and $y_{n_3}=2$. Then $z_{n_3}=1$ is the right choice:
$$\vec n_3=\begin{bmatrix} 0\\ 2\\ 1\end{bmatrix}$$
will be perpendicular to $\vec d$.
In order to get the equation of the third plane we need a point on the intersection line. For example, a point belonging to $t=0$, a common point of the two planes given satisfies that requirement. So the normal vector is $\vec n_3$ and a point on the third plane is $(1,1,0).$
Let $(x,y,z)$ be an arbitrary point of the third plane. Then the vector
\begin{bmatrix} x-1\\ y-1\\ z \end{bmatrix}
will be parallel to it and the scalar product of this vector and $n_3$ will be zero:
$$\vec d \cdot \begin{bmatrix} x-1\\ y-1\\ z \end{bmatrix}= \begin{bmatrix} 0\\ 2\\ 1\end{bmatrix}\cdot \begin{bmatrix} x-1\\ y-1\\ z \end{bmatrix}=0.$$
Hence, the point-normal equation of a suitable third plane is
$$2(y-1)+z=2y+z-2=0.$$
Let's check if the intersection line of the first two plane is in the third one. The equation of the intersection line is
$$\begin{bmatrix} x(t)\\ y(t)\\ z(t)\end{bmatrix}=\begin{bmatrix} 5\\ 1\\ -2\end{bmatrix}t+\begin{bmatrix} 1\\ 1\\ 0\end{bmatrix}.$$
Substituting $x(t)$, $y(t)$, and $z(t)$ into the equation of the third plane will result in $0$. Finally we can see that $n_3$ is not a scalar multiple of either $n_1$ or $n_2$.
The following figure illustrates what we have been doing:
Best Answer
Ask yourself what information you need in order to produce an equation of a plane, e.g., three noncolinear points on the plane, a point on the plane and a normal vector, &c. Then, look at the information you’ve been given and try to figure out how to get the data that you need from the given data. For instance, from the line that must lie on the plane you can get two of the three noncolinear points that you need; from two lines that are parallel to the plane (and are not themselves parallel) you can get its normal, and so on.