Lines and Planes

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Objectives:

1.: Be able to determine the equation of a line in 2D or 3D (the vector equation or the symmetric equations)
2.: Be able to determine whether two lines intersect or are skew.
3.: Know the equation of a plane.
4.: Be able to work various problems related to lines and planes.

Recap Video

You can watch watch the following video which recaps the ideas of the section.

Test your understanding with the following questions.

Which of the following is an example of a vector equation for a line?

$\textbf {r}(t) = \left \langle 1+t,2+t^2,3-t \right \rangle$ $\textbf {r}(t) = \left \langle 1+t,1+t,e^t \right \rangle$ $\textbf {r}(t) = \left \langle 1+t,1+t, 3+2t \right \rangle$

What two pieces of information do you need to write down the equation of a plane?

A point on the plane and a vector perpendicular to the plane. A point on the plane and a vector lying in the plane. A vector lying in the plane and a vector perpendicular to the plane.

Line Example

Below is a video showing an example of finding the vector equation and symmetric equations for a line.

Find the vector equation and symmetric equations for the line passing through $P(1,2,1)$ and $Q(-3,5,1)$ .

Line Problems

Consider the line $L$ given by the vector equation $\textbf {r}(t) = \left \langle 3t+2,2t+3,1 \right \rangle$ . Is the point $(5,5,1)$ on the line?

Yes No

See if you can find a value of $t$ that gives $(5,5,1)$ as the output.

Find the vector equation of the line passing through the point $P(1,2,1)$ and $Q(2,1,0)$ .

Steps:

What is the direction vector from $P$ to $Q$ ? $\textbf {v} = \left \langle \answer {1},\answer {-1},\answer {-1} \right \rangle$
If $P$ is the choice of initial point for the line, then the vector equation is $\textbf {r}(t) = \left \langle \answer {1},\answer {2},\answer {1} \right \rangle + t \left \langle \answer {1},\answer {-1},\answer {-1} \right \rangle .$

Let $L_1$ be the line through $(0,-1,1)$ and $(2,2,3)$ , and $L_2$ the line given by $\textbf {r}_2(t) = \left \langle 4t,5t,5t \right \rangle$ .

A vector equation of the line $L_1$ is $\textbf {r}_1(t) = \left \langle \answer {0},\answer {-1},1 \right \rangle + t\left \langle 2,\answer {3},\answer {2} \right \rangle .$
Do the lines $L_1$ and $L_2$ intersect?
Steps:
We need to set $\textbf {r}_1(t) = \textbf {r}_2(s)$ . When we do this we get three equations: $(1): 2t =\answer {4s}$ $(2): 3t-1 = \answer {5s}$ $(3): 2t+1 = \answer {5s}$
Solving the first two equations for $s$ and $t$ gives: $t=\answer {2}$ and $s = \answer {1}$
Do your values for $s$ and $t$ work in the third equation? Use this to decide: do the lines intersect?
Yes No

The point of intersection is $(\answer {4},\answer {5},\answer {5})$ .

The following problem will show how nice vectors can be:

What is the angle between the lines in $\mathbb {R}^2$ given by $L_1:y = 2x+1$ and $L_2:y=-x$ ?

Steps:

The direction vector for the line $L_1$ with $x$ -component $1$ is: $\left \langle 1, \answer {2} \right \rangle$ .
The direction vector for the line $L_2$ with $x$ -component $1$ is: $\left \langle 1,\answer {-1} \right \rangle$
The angle between the lines is the angle between the direction vectors. What is the angle between the direction vectors? $\cos ^{-1}\left (\answer {-1/\sqrt {10}} \right )$ .

Plane Problems

A normal vector for the plane $-x+2y=5$ is $\left \langle -1,\answer {2},\answer {0}\right \rangle .$

The equation of the plane passing through $P(2,1,-1)$ , $Q(0,-2,0)$ , and $R(1,-1,2)$ is:

Steps:

The vector $\overrightarrow {PQ} = \left \langle \answer {-2},\answer {-3},\answer {1} \right \rangle$ and $\overrightarrow {PR} = \left \langle \answer {-1},\answer {-2},\answer {3} \right \rangle$ .
The cross product is $\overrightarrow {PR} \times \overrightarrow {PQ} =\left \langle \answer {7},\answer {-5},\answer {-1} \right \rangle$ .
If we use $P$ in the equation of the plane, we get $7(x-2)+\answer {-5}(y-\answer {1})+\answer {-1}(z-\answer {-1}) = 0.$

The point of intersection between the line $\textbf {r}(t) = \left \langle 1-3t,2t,t-1 \right \rangle$ and the plane $x+y+2z=0$ is: $(\answer {-2},\answer {2},\answer {0})$ .

Plug the parametric equations of the line into the plane equation.

Find the angle between the planes $P_1: x-y+z=1$ and $P_2: y+z=3$ .

Steps:

A normal vector for $P_1$ is the vector $\textbf {n}_1 = \left \langle 1,\answer {-1},\answer {1} \right \rangle$ and a normal vector for $P_2$ is $\textbf {n}_2 = \left \langle \answer {0},1,\answer {1} \right \rangle$ .
The dot product of $\textbf {n}_1$ and $\textbf {n}_2$ is $\textbf {n}_1 \cdot \textbf {n}_2 = \answer {0}$
The angle between the planes is $\theta =\answer {\pi /2}$ .

True/False

Throughout, $\textbf {0}$ will denote the zero vector.

The line of intersection of the planes $x-y+z=1$ and $y+z=3$ is parallel to $\textbf {v} = \left \langle 1,-1,1 \right \rangle$ . TrueFalse

The equation $\textbf {r}(t) = \left \langle 1+2t,2+3t,1-4t \right \rangle$ is a vector equation for the line through $P(1,2,1)$ and perpendicular to the plane $2x+3y-4z=10$ . TrueFalse

Optional: Symmetric Equations Explained

For those who are interested, below is a video explaining the geometric significance of the symmetric equations for a line from the book.