Lines and Planes in Three Dimensions

Parametric Equations of a Line

The parametric equations of a line in R3\mathbb{R}^3 are a natural extension of the ones for a line in R2\mathbb{R}^2. A line passing through (x0,y0,z0)(x_0,y_0,z_0) going in the direction of the vector a,b,c⟨a,b,c⟩ is expressed by the following set of parametric equations: x=x0+at        <t<y=y0+btz=z0+ct\begin{aligned} x &= x_0+at \ \ \ \ \ \ \ \ -\infty < t < \infty\ y &= y_0+bt\ z &= z_0+ct \end{aligned}

Find the equation of the line through (2,1,3)(2,−1,3) and parallel to the vector 2i^+j^5k^2\hat{i}+\hat{j}−5\hat{k}.

Solution

The initial point is (x0,y0,z0)=(2,1,3)(x_0,y_0,z_0)=(2,−1,3) and the direction is 2,1,5⟨2,1,−5⟩, so the parametric equations for the line are x=2+2t        <t<y=1+tz=35t\ans{\begin{aligned} x &= 2+2t \ \ \ \ \ \ \ \ -\infty < t < \infty\ y &= -1+t\ z &= 3-5t \end{aligned}}

These parametric equations can also be written in vector form: l=x0,y0,z0+ta,b,c\vec{l} = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle l=2,1,3+t2,1,5\ans{\vec{l}=\langle 2,-1,3 \rangle + t\langle 2,-1,5 \rangle}

Symmetric Form

This line can also be expressed by solving for t: t=xx0a=yy0b=zz0ct=\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}

For example, the line in the previous example can be written x22=y+11=z35\dfrac{x-2}{2} = \dfrac{y+1}{1} = \dfrac{z-3}{-5}

  1. Find the line that passes through the points (3,5,8)(−3,5,8) and (4,2,1)(4,2,−1).

  2. l=3,5,8+t7,3,9\vec{l} = \langle -3,5,8 \rangle + t\langle 7,-3,-9 \rangle

Find the projection of the line l=3,5,8+t7,3,9\vec{l} =⟨−3,5,8⟩+t⟨7,−3,−9⟩

onto the xyxy, xzxz, and yzyz planes.

Solution

  1. xyxy plane: z=0z=0 x=3+7ty=53t\begin{aligned} x &= -3+7t\ y &= 5-3t \end{aligned} Eliminating the parameter: y=37x+267\ans{y=-\dfrac{3}{7}x+\dfrac{26}{7}}

  2. xzxz plane: y=0y=0 x=3+7tz=89t\begin{aligned} x &= -3+7t\ z &= 8-9t \end{aligned} Eliminating the parameter: z=97x+297\ans{z=-\dfrac{9}{7}x+\dfrac{29}{7}}

  3. yzyz plane: x=0x=0 y=53tz=89t\begin{aligned} y &= 5-3t\ z &= 8-9t \end{aligned} Eliminating the parameter: z=3y7\ans{z=3y-7}

Equation of a Plane

A plane in R3\mathbb{R}^3 is a natural extension of a line in R2\mathbb{R}^2. For example, consider the line 2x+3y=62x+3y=6, which can be rewritten y=23x+2y=−\dfrac{2}{3}x+2. The "direction" of this line is the vector 2,3⟨2,3⟩, the coefficients of xx and yy in the equation. Also, lines that are perpendicular to this line have slope 32\dfrac{3}{2}, which points in the same direction as the direction vector (2,3⟨2,3⟩; remember rise over run).

The extension of this to planes in R3\mathbb{R}^3 goes like this: the plane has a direction vector a,b,c⟨a,b,c⟩ that is normal (perpendicular) to the plane.

Plane with normal vector a,b,c

The equation of a plane in R3\mathbb{R}^3 is given by ax+by+cz=d    or    a(xx0)+b(yy0)+c(zz0)=0ax+by+cz = d\ \ \ \textrm{ or }\ \ \ a(x-x_0)+b(y-y_0)+c(z-z_0) = 0 for some point (x0,y0,z0)(x_0,y_0,z_0) in the plane. The vector n=a,b,cn⃗ =⟨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)P(2,−1,3), Q(4,2,6)Q(4,2,6), and R(5,3,2)R(5,3,−2).

Solution

We need two things: a point on the plane and a vector normal to the plane.

  1. Point: (x0,y0,z0)=(2,1,3)(x_0,y_0,z_0)=(2,−1,3) (we could use any of the three given points)
  2. 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^j^k^233345=27i^+19j^k^\langle a,b,c \rangle = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ 2 & 3 & 3\ 3 & 4 & -5 \end{vmatrix} = -27\hat{i}+19\hat{j}-\hat{k}

Therefore, the plane can be written L:27(x2)+19(y+1)(z3)=0L:27x19y+z=76\begin{aligned} &\vec{L} : -27(x-2)+19(y+1)-(z-3)=0\ &\ans{\vec{L} : 27x-19y+z=76} \end{aligned}

  1. Find the equation of the plane that passes through the points P(2,1,3)P(2,−1,3), Q(1,4,0)Q(1,4,0), and R(0,1,5)R(0,−1,5).

  2. L:5x+4y+5z=21\vec{L} : 5x+4y+5z = 21

Where does the line x=2+3tx=2+3t, y=1+ty=−1+t, z=4tz=4t intersect the plane x+y+z=13x+y+z=13?

Solution

The plane and the line intersect where the xx, yy, and zz coordinates are equal. Therefore, plug the parametric equations into the equation of the plane: (2+3t)+(1+t)+(4t)=138t+1=13t=32\begin{aligned} (2+3t)+(-1+t)+(4t) &= 13\ 8t+1 = 13 &\longrightarrow t = \dfrac{3}{2} \end{aligned}

Plug this tt value into the parametric equations to find the coordinates of the intersection point: x=2+3(32)=132y=1+32=12z=4(32)=6\begin{aligned} x &= 2+3\left(\dfrac{3}{2}\right) = \dfrac{13}{2}\ y &= -1+\dfrac{3}{2} = \dfrac{1}{2}\ z &= 4\left(\dfrac{3}{2}\right) = 6 \end{aligned}

The intersection point, then, is (132,12,6)\ans{\left(\dfrac{13}{2},\dfrac{1}{2},6\right)}

  1. Where does the line x=1+2tx=−1+2t, y=3ty=3t, z=4tz=4−t intersect the plane 2x3yz=62x−3y−z=6?

  2. (1,0,4)(−1,0,4)

Find parametric equations for the line of intersection of the planes x+y2z=6x+y−2z=6 and 3x2y+4z=83x−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⟨1,1,−2⟩ and 3,2,4⟨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^j^k^112324=0,10,5\begin{aligned} \langle a,b,c \rangle &= \vec{n_1} \times \vec{n_2}\ &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\ 1 & 1 & -2\ 3 & -2 & 4 \end{vmatrix} = \langle 0,-10,-5 \rangle \end{aligned}

Therefore, the direction of the line is 0,10,5⟨0,−10,−5⟩, so all we need is a point on the line. x=x0y=y010tz=z05t\begin{aligned} x &= x_0\ y &= y_0 - 10t\ z &= z_0 - 5t \end{aligned}

To find (x0,y0,z0)(x_0,y_0,z_0), we'll let z=0z=0 in each of the two planes to find the point where this line intersects the xyxy plane.

If x+y=6x+y=6 and 3x2y=83x−2y=8, then x0=4x_0=4, y0=2y_0=2, and z0=0z_0=0, so the intersection line is given by the following parametric equations: x=4y=210tz=5t\ans{\begin{aligned} x &= 4\ y &= 2-10t\ z &= -5t \end{aligned}}

  1. Find parametric equations for the line of intersection of the planes x+y2z=6x+y−2z=6 and 3x2y+4z=83x−2y+4z=8.

  2. x=4y=210tz=5t\begin{aligned} x &= 4\ y &= 2-10t\ z &= -5t \end{aligned}

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+y2z=6x+y−2z=6 and 3x2y+4z=83x−2y+4z=8.

Solution

Find the dot product of the normal vectors: n1n2=7=n1n2cosθ7=(1+1+4)(9+4+16)cosθcosθ=76 29θ112.3\begin{aligned} \vec{n_1} \cdot \vec{n_2} = -7 &= |\vec{n_1}| |\vec{n_2}| \cos \theta\ \ -7 &= (\sqrt{1+1+4})(\sqrt{9+4+16}) \cos \theta\ \cos \theta &= \dfrac{-7}{\sqrt{6}\ \sqrt{29}}\ &\longrightarrow \ans{\theta \approx 112.3^{\circ}} \end{aligned}

  1. Find the angle between the planes 5x+4y+5z=215x+4y+5z=21 and 2x3yz=62x−3y−z=6.

  2. θ=102.9\theta = 102.9^{\circ}

Parallel and Orthogonal Planes

Which of the following planes are parallel and which are orthogonal? Q:2x3y+6z=12R:x+32y3z=14S:6x+8y+2z=1T:9x12y3z=7\begin{array}{l l} Q: 2x-3y+6z=12 & R: -x+ \dfrac{3}{2}y-3z=14\ S: 6x+8y+2z=1 & T: -9x-12y-3z=7 \end{array}

Solution

nQ:2,3,6nR:1,32,3nS:6,8,2nR:9,12,3\begin{array}{ll} \vec{n_Q}: \langle 2,-3,6 \rangle & \vec{n_R}: \left\langle -1,\dfrac{3}{2},-3 \right\rangle\ \vec{n_S}: \langle 6,8,2 \rangle & \vec{n_R}: \langle -9,-12,-3 \rangle \end{array}

  1. nQnR=29218=492\vec{n_Q} \cdot \vec{n_R} = -2-\dfrac{9}{2}-18 = -\dfrac{49}{2} nQnR=(1+94+9)(4+9+36)=492|\vec{n_Q}| |\vec{n_R}| = (\sqrt{1+\dfrac{9}{4}+9})(\sqrt{4+9+36}) = \dfrac{49}{2}

Therefore, cosθ=49/249/2=1θ=π\cos \theta = \dfrac{-49/2}{49/2} = -1 \longrightarrow \theta = \pi which means that the planes are parallel (note: nQ=2nR\vec{n_Q} = -2\vec{n_R}).

  1. nSnT=156\vec{n_S} \cdot \vec{n_T} = -156 nSnT=(36+64+4)(81+144+9)=156|\vec{n_S}| |\vec{n_T}| = (\sqrt{36+64+4})(\sqrt{81+144+9}) = 156

Therefore, cosθ=1θ=π\cos θ=−1⟶θ=π, which means that the planes are parallel (note: nT=32nS\vec{n_T}=−\dfrac{3}{2}\vec{n_S}).

  1. nQnS=0\vec{n_Q} \cdot \vec{n_S} = 0, so the planes are orthogonal.

Therefore, QRST\ans{Q \parallel R \perp S \parallel T}

  1. Are x+y+4z=10x+y+4z=10 and x3y+z=10−x−3y+z=10 parallel, orthogonal, or neither?

  2. Orthogonal

  3. Are 2x+2y3z=102x+2y−3z=10 and 10x10y+15z=10−10x−10y+15z=10 parallel, orthogonal, or neither?

  4. Parallel