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13.5 Lines and Planes

line: collection of points that lie along a certain direction

Figure: A dotted line with a green arrow indicating direction.

the direction vector (like slope in \(\mathbb{R}^2\))

to find the direction vector, find the position vector of one point from any other

if we know one point on the line: \((x_0, y_0, z_0)\)

3D coordinate system with x, y, z axes showing a line and vectors \vec{r}_0 and \vec{r}_0 + t\vec{v}. — Figure: 3D coordinate system with x, y, z axes showing a line and vectors \(\vec{r}_0\) and \(\vec{r}_0 + t\vec{v}\).

how to write the equation of this line \(\rightarrow\) finding the position vector of any point on this line

\(\vec{r}_0\) : vector from origin to \((x_0, y_0, z_0)\)
\(\vec{v}\) : direction vector

any other point is reached by

\[\vec{r}(t) = \vec{r}_0 + t \vec{v}\]

\(-\infty < t < \infty\)

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\(\vec{r}(t) = \vec{r}_0 + t \vec{v}\) is called the vector form of eq. of a line

if \(\vec{v} = \langle a, b, c \rangle\)

\(\vec{r}_0 = \langle x_0, y_0, z_0 \rangle\)

then

\[\vec{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle\]

\[= \langle x_0 + at, y_0 + bt, z_0 + ct \rangle\]

\(x(t) = x_0 + at\)

\(y(t) = y_0 + bt\)

\(z(t) = z_0 + ct\)

parametric form

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Example: Line through Points

Find the equation of a line through points P(0, 1, 2) and Q(-3, 4, 7).

Figure: A 3D Cartesian coordinate system with x, y, and z axes. A line passes through points P and Q. A vector r0 points from the origin to P.

Direction vector:Some multiple of \(\vec{PQ}\) or \(\vec{QP}\)
Let's do \(\vec{PQ}\):
\[ \vec{v} = \vec{PQ} = \langle -3, 3, 5 \rangle \]
\(\vec{r}_0\):From origin to any known point
Let's do \(\vec{r}_0 = \vec{OP}\):
\[ \vec{r}_0 = \langle 0, 1, 2 \rangle \]

Vector Form

Then \(\vec{r}(t) = \vec{r}_0 + t\vec{v}\)

\[ \begin{aligned} \vec{r}(t) &= \langle 0, 1, 2 \rangle + t \langle -3, 3, 5 \rangle \\ &= \langle -3t, 1 + 3t, 2 + 5t \rangle \end{aligned} \]

Parametric Form

\[ \begin{cases} x = -3t \\ y = 1 + 3t \\ z = 2 + 5t \end{cases} \]

\(-\infty < t < \infty\) — infinitely long line

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Restricting to a Line Segment

If we just want the segment from P to Q, restrict \(t\):

Notice if \(t = 0\), \(\vec{r}(0) = \langle 0, 1, 2 \rangle\) — tip is at P
Notice if \(t = 1\), \(\vec{r}(1) = \langle 0, 1, 2 \rangle + \langle -3, 3, 5 \rangle = \langle -3, 4, 7 \rangle\) — tip is at Q

So, \(0 \le t \le 1\) we get the segment \(\vec{PQ}\)

Plane:

\((x_0, y_0, z_0)\): known point
\((x, y, z)\): some other point
Vector from \((x_0, y_0, z_0)\) to \((x, y, z)\) is orthogonal to a vector perpendicular to the plane ("normal vector")

We know \(\vec{n} \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0\)

\[ \langle a, b, c \rangle \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0 \]

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\[ a(x-x_0) + b(y-y_0) + c(z-z_0) = 0 \]

eq. of plane through \( (x_0, y_0, z_0) \) with normal vector \( \vec{n} = \langle a, b, c \rangle \)

example

Plane containing the vectors \( \vec{u} = \langle 0, 1, 2 \rangle \), \( \vec{v} = \langle -1, -3, 0 \rangle \) and point \( (-4, 7, 5) \)

Figure: Diagram of a plane in 3D space containing vectors u and v, with normal vectors n = u x v and n = v x u.

\( \vec{n} \) is perpendicular to BOTH \( \vec{u}, \vec{v} \)

so, \( \vec{n} = \vec{u} \times \vec{v} \) or \( \vec{v} \times \vec{u} \)

let's do \( \vec{n} = \vec{u} \times \vec{v} \)

\( = \dots = \langle 6, -2, 1 \rangle \)

\( \vec{n} \cdot \langle x+4, y-7, z-5 \rangle = 0 \)

\( \langle 6, -2, 1 \rangle \cdot \langle x+4, y-7, z-5 \rangle = 0 \)

\[ 6(x+4) - 2(y-7) + (z-5) = 0 \]

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in \( \mathbb{R}^2 \), if two lines do not have same slope, they must intersect

Figure: 2D graph with x and y axes showing two lines intersecting at a single point.

that is NOT true in \( \mathbb{R}^3 \) (or higher)

e.g. \( \vec{u} = \langle 1, 0, 1 \rangle + t \langle 1, 2, 3 \rangle \)

\( \vec{v} = t \langle 1, 0, 0 \rangle \)

Figure: 3D coordinate system with x, y, and z axes showing two non-parallel, non-intersecting skew lines.

these lines never intersect

(gap in \( z \) direction)

we say these lines are skew

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Relationships Between Planes

Similarly, two planes are parallel if their normal vectors are parallel.

Two planes are orthogonal if their normal vectors are orthogonal.

Figure: Two parallel planes depicted as parallelograms with parallel normal vectors labeled n.

Figure: Two perpendicular planes with normal vectors that are also perpendicular to each other.

Intersection of Two Planes

The intersection of two planes is a line.

Figure: Two intersecting planes showing the line of intersection and normal vectors n1 and n2.

\(\vec{v}\), the direction vector of the line of intersection is \(\perp\) to both of the normal vectors.

\[ \vec{v} = \vec{n}_1 \times \vec{n}_2 \quad \text{or} \quad \vec{n}_2 \times \vec{n}_1 \]

The line of intersection contains pts that are on both planes.

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Finding Normal Vectors from Equations

If given a plane equation, normal vectors can be found by looking at coefficients of \(x, y, z\).

6\((x + 4)\) -2\((y - 7)\) + 1\((z - 5) = 0\)

\(\vec{n} = \langle 6, -2, 1 \rangle\) or \(\langle -6, 2, -1 \rangle\) or any multiple of either.

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How to find a point on the line of intersection of two planes?

extra

plane Q: \( x + 3y - 2z = 1 \)

plane R: \( x + y + z = 0 \)

Figure: A 3D sketch showing two intersecting planes, labeled Q and P, with a point (x,y,z) on their line of intersection.

we need \( x, y, z \) such that \( x + 3y - 2z = 1 \) AND \( x + y + z = 0 \) are both true.

we have 3 variables \( (x, y, z) \) and 2 constraints, so one variable is arbitrary (we can choose whatever we want).

as an example, I choose \( z = 0 \)

to find \( x, y \), I solve \( x + 3y - 2z = 1 \) and \( x + y + z = 0 \) simultaneously.

\[ x + 3y - 2(0) = 1 \quad \text{and} \quad x + y + (0) = 0 \]

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example

Supplemental example

Two objects travel on the lines

\[ \vec{r}_1(t) = \langle 2t+3, 4t+2, 3t+5 \rangle \quad -\infty < t < \infty \]\[ \vec{r}_2(s) = \langle s+2, 3s-1, -5s+10 \rangle \quad -\infty < s < \infty \]

will the objects' paths intersect?

will the objects collide with each other?

intersect: can we find \( t, s \) such that \( \vec{r}_1(t) = \vec{r}_2(s) \)?

collide: can we find \( t, s \) such that \( \vec{r}_1(t) = \vec{r}_2(s) \) AND \( t = s \)?

if \( \vec{r}_1(t) = \vec{r}_2(s) \) then

\[ x: 2t+3 = s+2 \quad \text{--- (1)} \]\[ y: 4t+2 = 3s-1 \quad \text{--- (2)} \]\[ z: 3t+5 = -5s+10 \quad \text{--- (3)} \]

from (1), \( s = 2t + 1 \)

sub into (2): \( 4t + 2 = 3(2t + 1) - 1 \)

\( t = 0 \) so, \( s = 1 \)

Check if these work in (3)

(3): \( 3(0) + 5 = -5(1) + 10 \)?

yes, so at \( t=0, s=1 \), \( \vec{r}_1(t) = \vec{r}_2(s) \) they intersect