The parametric equations of a line in R3 are a natural extension of the ones for a line in R2. A line passing through (x0,y0,z0) going in the direction of the vector ⟨a,b,c⟩ is expressed by the following set of parametric equations:
xyz=x0+at−∞<t<∞=y0+bt=z0+ct
Find the equation of the line through (2,−1,3) and parallel to the vector 2i^+j^−5k^.
Solution
The initial point is (x0,y0,z0)=(2,−1,3) and the direction is ⟨2,1,−5⟩, so the parametric equations for the line are
xyz=2+2t−∞<t<∞=−1+t=3−5t
These parametric equations can also be written in vector form:
l=⟨x0,y0,z0⟩+t⟨a,b,c⟩l=⟨2,−1,3⟩+t⟨2,−1,5⟩
Symmetric Form
This line can also be expressed by solving for t:
t=ax−x0=by−y0=cz−z0
For example, the line in the previous example can be written
2x−2=1y+1=−5z−3
Find the line that passes through the points (−3,5,8) and (4,2,−1).
l=⟨−3,5,8⟩+t⟨7,−3,−9⟩
Find the projection of the line l=⟨−3,5,8⟩+t⟨7,−3,−9⟩
onto the xy, xz, and yz planes.
Solution
xy plane: z=0xy=−3+7t=5−3t
Eliminating the parameter:
y=−73x+726
xz plane: y=0xz=−3+7t=8−9t
Eliminating the parameter:
z=−79x+729
yz plane: x=0yz=5−3t=8−9t
Eliminating the parameter:
z=3y−7
Equation of a Plane
A plane in R3 is a natural extension of a line in R2. For example, consider the line 2x+3y=6, which can be rewritten y=−32x+2. The "direction" of this line is the vector ⟨2,3⟩, the coefficients of x and y in the equation. Also, lines that are perpendicular to this line have slope 23, which points in the same direction as the direction vector (⟨2,3⟩; remember rise over run).
The extension of this to planes in R3 goes like this: the plane has a direction vector ⟨a,b,c⟩ that is normal (perpendicular) to the plane.
The equation of a plane in R3 is given by
ax+by+cz=d or a(x−x0)+b(y−y0)+c(z−z0)=0
for some point (x0,y0,z0) in the plane. The vector n⃗=⟨a,b,c⟩ is any normal vector to the plane (this is what specifies the "slope" of the plane).
Find the equation of the plane through the points P(2,−1,3), Q(4,2,6), and R(5,3,−2).
Solution
We need two things: a point on the plane and a vector normal to the plane.
Point: (x0,y0,z0)=(2,−1,3) (we could use any of the three given points)
Normal vector: take the cross product of two vectors in the plane (vectors that connect two of the points given)
⟨a,b,c⟩=PQ×PR=∣∣∣∣∣∣i^23j^34k^3−5∣∣∣∣∣∣=−27i^+19j^−k^
Therefore, the plane can be written
L:−27(x−2)+19(y+1)−(z−3)=0L:27x−19y+z=76
Find the equation of the plane that passes through the points P(2,−1,3), Q(1,4,0), and R(0,−1,5).
L:5x+4y+5z=21
Where does the line x=2+3t, y=−1+t, z=4t intersect the plane x+y+z=13?
Solution
The plane and the line intersect where the x, y, and z coordinates are equal. Therefore, plug the parametric equations into the equation of the plane:
(2+3t)+(−1+t)+(4t)8t+1=13=13⟶t=23
Plug this t value into the parametric equations to find the coordinates of the intersection point:
xyz=2+3(23)=213=−1+23=21=4(23)=6
The intersection point, then, is
(213,21,6)
Where does the line x=−1+2t, y=3t, z=4−t intersect the plane 2x−3y−z=6?
(−1,0,4)
Find parametric equations for the line of intersection of the planes x+y−2z=6 and 3x−2y+4z=8.
Solution
The intersection line will lie in both planes at the same time, and thus will be perpendicular to the normal vectors for both planes. The normal planes are ⟨1,1,−2⟩ and ⟨3,−2,4⟩. To find the direction of the line (perpendicular to these two vectors), take the cross product of these normal vectors.
⟨a,b,c⟩=n1×n2=∣∣∣∣∣∣i^13j^1−2k^−24∣∣∣∣∣∣=⟨0,−10,−5⟩
Therefore, the direction of the line is ⟨0,−10,−5⟩, so all we need is a point on the line.
xyz=x0=y0−10t=z0−5t
To find (x0,y0,z0), we'll let z=0 in each of the two planes to find the point where this line intersects the xy plane.
If x+y=6 and 3x−2y=8, then x0=4, y0=2, and z0=0, so the intersection line is given by the following parametric equations:
xyz=4=2−10t=−5t
Find parametric equations for the line of intersection of the planes x+y−2z=6 and 3x−2y+4z=8.
xyz=4=2−10t=−5t
The Angle Between Two Planes
This is pretty straightforward: the angle between two planes is the same as the angle between their normal vectors. And remember, we can find the angle between two vectors using their dot product.
Find the angle between the planes x+y−2z=6 and 3x−2y+4z=8.
Solution
Find the dot product of the normal vectors:
n1⋅n2=−7−7cosθ=∣n1∣∣n2∣cosθ=(1+1+4)(9+4+16)cosθ=629−7⟶θ≈112.3∘
Find the angle between the planes 5x+4y+5z=21 and 2x−3y−z=6.
θ=102.9∘
Parallel and Orthogonal Planes
Parallel: their normal vectors are parallel (θ=0 or π)
Orthogonal: their normal vectors are orthogonal (n1⋅n1=0)
Which of the following planes are parallel and which are orthogonal?
Q:2x−3y+6z=12S:6x+8y+2z=1R:−x+23y−3z=14T:−9x−12y−3z=7