The parametric equations of a line in \(\mathbb{R}^3\) are a natural extension of the ones for a line in \(\mathbb{R}^2\). A line passing through \((x_0,y_0,z_0)\) going in the direction of the vector \(\langle a,b,c \rangle\) is expressed by the following set of parametric equations:

\[\begin{align} x &= x_0+at \ \ \ \ \ \ \ \ -\infty < t < \infty\\ y &= y_0+bt\\ z &= z_0+ct \end{align}\]Find the equation of the line through \((2,-1,3)\) and parallel to the vector \(2\hat{i}+\hat{j}-5\hat{k}\).

The initial point is \((x_0,y_0,z_0) = (2,-1,3)\) and the direction is \(\langle 2,1,-5 \rangle\), so the parametric equations for the line are

\[\ans{\begin{align} x &= 2+2t \ \ \ \ \ \ \ \ -\infty < t < \infty\\ y &= -1+t\\ z &= 3-5t \end{align}}\]These parametric equations can also be written in vector form:

\[\vec{l} = \langle x_0,y_0,z_0 \rangle + t\langle a,b,c \rangle\] \[\ans{\vec{l}=\langle 2,-1,3 \rangle + t\langle 2,-1,5 \rangle}\]This line can also be expressed by solving for t:

\[t=\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}\]For example, the line in the previous example can be written

\[\dfrac{x-2}{2} = \dfrac{y+1}{1} = \dfrac{z-3}{-5}\]Find the line that passes through the points \((-3,5,8)\) and \((4,2,-1)\).

\(\vec{l} = \langle -3,5,8 \rangle + t\langle 7,-3,-9 \rangle\)

Find the projection of the line \(\vec{l} = \langle -3,5,8 \rangle + t\langle 7,-3,-9 \rangle\) onto the xy, xz, and yz planes.

- xy plane: \(z=0\) \[\begin{align} x &= -3+7t\\ y &= 5-3t \end{align}\] Eliminating the parameter: \[\ans{y=-\dfrac{3}{7}x+\dfrac{26}{7}}\]
- xz plane: \(y=0\) \[\begin{align} x &= -3+7t\\ z &= 8-9t \end{align}\] Eliminating the parameter: \[\ans{z=-\dfrac{9}{7}x+\dfrac{29}{7}}\]
- yz plane: \(x=0\) \[\begin{align} y &= 5-3t\\ z &= 8-9t \end{align}\] Eliminating the parameter: \[\ans{z=3y-7}\]

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

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

The equation of a plane in \(\mathbb{R}^3\) is given by \[ax+by+cz = d\ \ \ \textrm{ or }\ \ \ a(x-x_0)+b(y-y_0)+c(z-z_0) = 0\] for some point \((x_0,y_0,z_0)\) in the plane. The vector \(\vec{n}=\langle a,b,c \rangle\) 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 \(R(5,3,-2)\).

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

- Point: \((x_0,y_0,z_0)=(2,-1,3)\) (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) \[\langle a,b,c \rangle = \vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 2 & 3 & 3\\ 3 & 4 & -5 \end{vmatrix} = -27\hat{i}+19\hat{j}-\hat{k}\]

Therefore, the plane can be written

\[\begin{align} &\vec{L} : -27(x-2)+19(y+1)-(z-3)=0\\ &\ans{\vec{L} : 27x-19y+z=76} \end{align}\]Find the equation of the plane that passes through the points \(P(2,-1,3)\), \(Q(1,4,0)\), and \(R(0,-1,5)\).

\(\vec{L} : 5x+4y+5z = 21\)

Where does the line \(x=2+3t\), \(y=-1+t\), \(z=4t\) intersect the plane \(x+y+z=13\)?

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:

\[\begin{align} (2+3t)+(-1+t)+(4t) &= 13\\ 8t+1 = 13 &\longrightarrow t = \dfrac{3}{2} \end{align}\]Plug this t value into the parametric equations to find the coordinates of the intersection point:

\[\begin{align} x &= 2+3\left(\dfrac{3}{2}\right) = \dfrac{13}{2}\\ y &= -1+\dfrac{3}{2} = \dfrac{1}{2}\\ z &= 4\left(\dfrac{3}{2}\right) = 6 \end{align}\]The intersection point, then, is

\[\ans{\left(\dfrac{13}{2},\dfrac{1}{2},6\right)}\]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 \(3x-2y+4z=8\).

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 \(\langle 1,1,-2 \rangle\) and \(\langle 3,-2,4 \rangle\). To find the direction of the line (perpendicular to these two vectors), take the cross product of these normal vectors.

\[\begin{align} \langle a,b,c \rangle &= \vec{n_1} \times \vec{n_2}\\ &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 1 & -2\\ 3 & -2 & 4 \end{vmatrix} = \langle 0,-10,-5 \rangle \end{align}\]Therefore, the direction of the line is \(\langle 0,-10,-5 \rangle\), so all we need is a point on the line. \[\begin{align} x &= x_0\\ y &= y_0 - 10t\\ z &= z_0 - 5t \end{align}\] To find \((x_0,y_0,z_0)\), 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 \(x_0=4\), \(y_0=2\), and \(z_0=0\), so the intersection line is given by the following parametric equations:

\[\ans{\begin{align} x &= 4\\ y &= 2-10t\\ z &= -5t \end{align}}\]Find parametric equations for the line of intersection of the planes \(x+y-2z=6\) and \(3x-2y+4z=8\).

\[\begin{align}
x &= 4\\
y &= 2-10t\\
z &= -5t
\end{align}\]

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 \(3x-2y+4z=8\).

Find the dot product of the normal vectors:

\[\begin{align} \vec{n_1} \cdot \vec{n_2} = -7 &= |\vec{n_1}| |\vec{n_2}| \cos \theta\\ \\ -7 &= (\sqrt{1+1+4})(\sqrt{9+4+16}) \cos \theta\\ \cos \theta &= \dfrac{-7}{\sqrt{6}\ \sqrt{29}}\\ &\longrightarrow \ans{\theta \approx 112.3^{\circ}} \end{align}\]Find the angle between the planes \(5x+4y+5z=21\) and \(2x-3y-z=6\).

\(\theta = 102.9^{\circ}\)

- Parallel: their normal vectors are parallel (\(\theta = 0\) or \(\pi\))
- Orthogonal: their normal vectors are orthogonal (\(\vec{n_1} \cdot \vec{n_1} = 0\))

Which of the following planes are parallel and which are orthogonal?

\[\begin{array} Q: 2x-3y+6z=12 & R: -x+ \dfrac{3}{2}y-3z=14\\ S: 6x+8y+2z=1 & T: -9x-12y-3z=7 \end{array}\]- \(\vec{n_Q} \cdot \vec{n_R} = -2-\dfrac{9}{2}-18 = -\dfrac{49}{2}\) \[|\vec{n_Q}| |\vec{n_R}| = (\sqrt{1+\dfrac{9}{4}+9})(\sqrt{4+9+36}) = \dfrac{49}{2}\] Therefore, \(\cos \theta = \dfrac{-49/2}{49/2} = -1 \longrightarrow \theta = \pi\), which means that the planes are parallel (note: \(\vec{n_Q} = -2\vec{n_R}\)).
- \(\vec{n_S} \cdot \vec{n_T} = -156\) \[|\vec{n_S}| |\vec{n_T}| = (\sqrt{36+64+4})(\sqrt{81+144+9}) = 156\] Therefore, \(\cos \theta = -1 \longrightarrow \theta = \pi\), which means that the planes are parallel (note: \(\vec{n_T} = -\dfrac{3}{2}\vec{n_S}\)).
- \(\vec{n_Q} \cdot \vec{n_S} = 0\), so the planes are orthogonal.

Therefore,

\[\ans{Q \parallel R \perp S \parallel T}\]