243.12.5.26 Show That The Line And The Plane Are Not Parallel.

  • Context: MHB 
  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Line Parallel Plane
Click For Summary
SUMMARY

The discussion focuses on demonstrating that the line defined by the vector equation \( r = r_0 + tv \) with \( r_0 = \langle 1, 1, 1 \rangle \) and \( v = \langle 1, 3, -1 \rangle \) is not parallel to the plane described by the equation \( x + 2y + 2z + 2 = 0 \). The analysis shows that the line intersects the plane at a specific point, confirming they are not parallel. The intersection occurs at \( t = -\frac{7}{5} \), leading to the coordinates of the intersection point being calculable by substituting this value back into the line's equations.

PREREQUISITES
  • Understanding of vector equations in three-dimensional space
  • Knowledge of plane equations and their representations
  • Familiarity with solving linear equations
  • Basic concepts of intersection in geometry
NEXT STEPS
  • Study the properties of lines and planes in three-dimensional geometry
  • Learn how to derive intersection points between lines and planes
  • Explore the concept of skew lines and their characteristics
  • Investigate the implications of parallel lines and planes in vector calculus
USEFUL FOR

Students and professionals in mathematics, particularly those studying geometry and vector calculus, as well as educators looking for examples of line-plane interactions.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$\textsf{Write a complete solution.}\\$
$\textit{Let}$ $$v =\langle 1, 3, − 1 \rangle$$
$\textit{and }$ $$r_0 =\langle 1, 1, 1 \rangle$$
$\textit{and consider the line given by:}\\$ $$r = r_0+tv$$
$\textit{in vector form.}\\$
$\textit{Also, consider the plane given by}$
$$x+2y+2z+2 = 0$$
$\textit{(a) Show that the line and the plane are not parallel.}\\$
$\textit{(b) Find the point on the line at distance 3 from the plane.}\\$

ok just posting this now to come back later to finish it.
to start with...
\begin{align*}\displaystyle
r&= r_0+tv\\
&=\langle 1, 1, 1 \rangle + t\langle 1, 3, − 1 \rangle\\
&=t+1,3t+1,-t+1
\end{align*}
 
Physics news on Phys.org
If a line and plane are not parallel then there must be a point where the line intersects the plane. (Note: two lines in 3 dimensions can intersect, be parallel, or be "skew", neither parallel nor intersecting. That cannot be true of two planes or a plane and a line. They must be parallel or intersecting.)

Here the line is given by x= 1+ t, y= 1+ 3t, and z= 1- t and the plane by x+ 2y+ 2z= -2. So, at a point of intersection, (1+t)+2(1+ 3t)+ 2(1- t)= 1+ t+ 2+ 6t+ 2- 2t= 5t+ 5= -2 so 5t= -7 and t= -7/5. Since there is such a t, there is a point intersection and you can find that point by putting that value of t into the equations of the line.
 

Similar threads

Replies
8
Views
3K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K