show that the line and plane are parallel..

Question

its a really stupid question but i cant seem to solve it. a line1 is defined by $2$ equation, $2x-z=7$ and $2y+z=9$ and a plane of $3x-y-2z=18$ is given. Now the question is to show that the line is parallel to the plane. I know the concept that for them to be parallel the dot product of normal of plane and the vector needs to be zero. But i dont understand the equations given for the line!! how do i know the vector ? please please help me out here thanks!!

Answer 1

This is a good question. Drawing a picture may also help.

1. Every plane can be written in the form $\vec{n}\cdot \vec{x} = C$, where $\vec{n}$ is a constant vector that is perpendicular to the plane, $C$ is a number, and $\vec{x}$ is an arbitrary point $\langle x, y, z\rangle$. If $\vec{x}$ is a point that satisfies this equation, it's on the plane. Otherwise, it's not on the plane.

2. If we can rewrite your plane in this form, we'll be able to read off $\vec{n}$, the vector perpendicular to the plane. But indeed we can rewrite $$3x - y - 2z = 18$$ as

   $$\begin{bmatrix}3&-1&-2\end{bmatrix}\cdot \begin{bmatrix}x\\y\\z \end{bmatrix} = 18$$

   So, the vector $\vec{n} \equiv [3, -1, -2]$ is perpendicular to the plane.

3. Now the line is defined by two equations: $2x-z=7$ and $2y+z=9$ (these are planes; their intersection gives the line).

   By the reasoning above, these two planes have normal vectors $\langle 2, 0, -1\rangle$ and $\langle 0, 2, 1\rangle$ respectively. If you envision two planes intersecting at a line in space, note that the line is parallel to both planes and therefore perpendicular to both normals.

4. To find a vector that is parallel to the line and perpendicular to both normals, we can take the cross product of both normals.

   $$\langle 2, 0, -1\rangle \times \langle 0, 2, 1\rangle = \langle 2, -2, 4\rangle$$.

5. Note that to show that the line is parallel to the plane, all we need to do is show that this vector $\langle 2, -2, 4\rangle$ is perpendicular to $\vec{n} = \langle 3, -1, -2\rangle$.

6. Note that $\langle 2, -2, 4\rangle \cdot \langle 3, -1, -2\rangle = 6 + 2 -8 = 0$, which establishes what we wanted to show.

Answer 2

Convert your line into a parametric equation of a line.

You need $2$ points.

1) Add equations:

$2x+2y =16$ , or $x+y =8;$

2)$ z=2x-7;$

Points: $( 4,4,1)$, $(5,3,3)$, for example.

$\vec r = (4,4,1) + t (1,-1,2)$;

where the direction vector:

$ \vec d =$

$ (5-4,3-4,3-1) =(1,-1,2)$

the vector pointing from

$(4, 4,1)$ to $(5,3,3).$

Show the line is parallel to the plane.

Normal vector of the plane :

$\vec n =(3,-1,-2)$.

Direction vector: $\vec d = (1,-1,2).$

Scalar product:

$\vec n \cdot \vec d= 3 +1 -4 =0.$

Hence line parallel to plane.