For questions involving lines and planes, we are usually asked to find the point of intersection. However, when we consider a line and a plane, there are three possible situations. In this page, we will consider all the possible outcomes.
1) the line intersects the plane
2) the line is parallel to the plane
3) the line lies in the plane
The following videos will help you understand all the concepts from this page
In the following video, we consider the three different possibilities
Example 1 - Point of Intersection
Find the point of intersection of the line \(-x=\frac {y-5}{2}=2z-8\) with the plane 3x – y + z = 8
Example 2 - No Intersection
Find the point of intersection of the line \(\textbf{r}=\left( \begin{matrix} 2 \\ 1\\0 \end{matrix} \right) +\mu\left( \begin{matrix} 4 \\ 3 \\1\end{matrix} \right) \)with the plane x - 2y + 2z = 6
Example 3 - Line lies in Plane
Find the point of intersection of the line \(x=2+4λ\ ,\ y=1+3λ\ ,\ z=λ\) with the plane x - 2y + 2z = 0
Notes from the video
In the following video, we look at a specific application of finding the intersection of a line with a plane. We will see how it is possible to find the coordinates of the point reflected in a plane. In order to do this, we have to find the equation of a line that goes through our point and is perpendicular to the plane.
Here is the example:
The point A (1, 3, 0) is on the line L, which is perpendicular to the plane \(3x−3y+2z=5\).
- Find the equation of the line L.
- Find the point R which is the intersection of the line L and the plane.
- The point A is reflected in the plane. Find the coordinates of the image of A.
Notes from the video
Print from here
Here is a quiz that practises the skills from this page
START QUIZ!
Find the value of \(\lambda\) in order that the point (6,1,-1) lies on the line \(\textbf{r}=\left( \begin{matrix} 2\\3 \\ -1 \end{matrix} \right) +\lambda \left( \begin{matrix} -2 \\ 1\\0 \end{matrix} \right) \)
\(\lambda\) =
\(\left( \begin{matrix} 2\\3 \\ -1 \end{matrix} \right) -2 \left( \begin{matrix} -2 \\ 1\\0 \end{matrix} \right) =\left( \begin{matrix} 6 \\ 1\\-1 \end{matrix} \right)\)
The point (2 , a , b ) lies on the line \(\textbf{r}=\left( \begin{matrix} -1\\0 \\ 2 \end{matrix} \right) +\lambda \left( \begin{matrix} 1 \\ -1\\2 \end{matrix} \right) \)
Find a and b
a =
b =
\(\left( \begin{matrix} -1\\0 \\ 2 \end{matrix} \right) +3 \left( \begin{matrix} 1 \\ -1\\2 \end{matrix} \right) =\left( \begin{matrix} 2 \\ a\\b \end{matrix} \right) \)
The point (1 , 2 , -1) lies in the plane x + 2y - 3z = a
Find a.
a =
x + 2y - 3z = a
(1) + 2(2) - 3(-1) = a
8 = a
The point (3 , 0 , k) lies in the plane 2x - y + 3z - 3 = 0
Find k.
k =
2x - y + 3z -3 = 0
2(3) - (0) + 3k - 3 = 0
3k = -3
k = -1
The line L does not intersect with the plane 4x + y + kz = 3 . Find k
L: \(x = 1+2\lambda\\ y = 3+\lambda\\ z=4-\lambda \)
k=
The line and plane must be parallel.
This means that the line is perpendicular to the normal.
The scalar product of the vectors equals zero.
\(\left( \begin{matrix} 2 \\ 1\\-1 \end{matrix} \right)\cdot \left( \begin{matrix} 4 \\ 1\\k \end{matrix} \right)=0\)
8 + 1 - k = 0
k = 9
(0 , a , b) is the point where the line line \(\textbf{r}=\left( \begin{matrix} 2\\1 \\ 4 \end{matrix} \right) +\lambda\left( \begin{matrix} -1 \\ 2\\3 \end{matrix} \right) \) meets the plane x = 0.
Find a and b.
a =
b =
\(\left( \begin{matrix} 2\\1 \\ 4 \end{matrix} \right) +\lambda\left( \begin{matrix} -1 \\ 2\\3 \end{matrix} \right) =\left( \begin{matrix} 0 \\ a\\b \end{matrix} \right) \)
\(\left( \begin{matrix} 2\\1 \\ 4 \end{matrix} \right) +2\left( \begin{matrix} -1 \\ 2\\3 \end{matrix} \right) =\left( \begin{matrix} 0 \\ a\\b \end{matrix} \right) \)
The line \(x=1+\lambda\\ y=1-\lambda\\ z=2+\lambda\) intersects with the plane x + y + z = 2 at the point (a , b , c).
Find a , b and c
a =
b =
c =
\(x + y + z = 2 \\ (1+\lambda)+(1-\lambda)+(2+\lambda)=2\\ 4+\lambda = 2\\ \lambda = -2\)
x = 1 + (-2)
y = 1 - (-2)
z = 2 + (-2)
The plane ax + 2y + 1z = 3 is perpendicular to the line \(\frac{x-1}{2}=y+1=2z\).
Find a
a =
If the plane and line are perpendicular, then the normal is parallel to the line.
The direction of the line is \(\left( \begin{matrix} 2 \\ 1\\0.5 \end{matrix} \right)\)
The direction of the normal is \(\left( \begin{matrix} a \\ 2\\1 \end{matrix} \right)\)
\(\left( \begin{matrix} a \\ 2\\1 \end{matrix} \right) =k\left( \begin{matrix} 2 \\ 1\\0.5 \end{matrix} \right)\)
\(\left( \begin{matrix} a \\ 2\\1 \end{matrix} \right) =2\left( \begin{matrix} 2 \\ 1\\0.5 \end{matrix} \right)\)
Find the intersection of the line L and the plane 3x - 2y + z = - 12 by filling in the blanks below.
\(L:\textbf{r}=\left( \begin{matrix} 1\\0 \\ -5 \end{matrix} \right) +\lambda\left( \begin{matrix} -2 \\ 1\\3 \end{matrix} \right) \)
\(x = 1 - 2\lambda\\ y = \lambda\\ z = -5+3\lambda\\\)
\(3(1 - 2\lambda)-2\lambda+(-5+3\lambda)=-12\)
\(3 - 6\lambda-2\lambda-5+3\lambda=-12\)
\(-5\lambda=\)
\(\lambda = \)
x =
y =
z =
The line L is perpendicular to the plane \(\Pi\)
The point of intersection of the line and the plane is (-2 , 1 , 2).
The point A(1 , 2 , 0) lies on the line L.
The reflection of the point A in the plane \(\Pi\) is (a , b , c)
Find a , b and c
a =
b =
c =
The points A and B are given by A(-8,1,-2) and B(-2,-1,2).
A plane Π is defined by the equation \(2x−y−3z=−8\)
- Find a vector equation of the line L passing through the points A and B.
- Find the coordinates of the point of intersection of the line and the plane.
Hint
Full Solution
Consider the plane \(x−2y+4z=−15\) and the line
\(x=3+kλ\\ y=−2+λ\\ z=(2k+6)−2λ\)
The line and the plane are perpendicular. Find
- The value of k
- The coordinates of the point of intersection of the line and the plane.
Hint
Full Solution
\(Π_{ 1 }\)and \(Π_{ 2 }\) are planes such that
\(Π_{ 1 }:2x−y−2z=0\)
and
\(Π_{ 2 }:−2x+3y+3z=4 \)
L is the intersection of planes \(Π_{ 1 }\) and \(Π_{ 2 }\)
- Find the equation of the line
A third plane \(Π_{ 3 }\) is defined by the equation \(kx+(k−1)y−z=5\)
- Find the value of k such that the line L does not intersect with \(Π_{ 3 }\)
Hint
Full Solution
The point A (3, 1, –2) is on the line L, which is perpendicular to the plane \(2x−3y−z+9=0\).
- Find the Cartesian equation of the line L.
- Find the point R which is the intersection of the line L and the plane.
- The point A is reflected in the plane. Find the coordinates of the image of A.
Hint
Full Solution
How much of Intersection of Line and Plane have you understood?
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