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# 1.6: Lines and Planes

### Lines

Our goal is to come up with the equation of a line given a vector v parallel to the line and a point (a,b,c) on the line. The figure (shown in 2D for simplicity) shows that if P is a point on the line then

$\langle x,y \rangle = P + tv$

for some number $$t$$.

The picture is the same for 3D.  The formula is given below.

Note: Parametric Equations of a Line

The parametric equations for the line through the point $$(a,b,c)$$ and parallel to the vector v are

$<x,y,z> = <a,b,c> + t\textbf{v}.$

Example 1

Find the parametric equations of the line that passes through the point $$(1, 2, 3)$$ and is parallel to the vector <4, -2, 1>.

Solution

We write:

$<x, y, z> = <1, 2, 3> + t <4, -2, 1> = <1 + 4t, 2 - 2t, 3 + t>$

or

$x(t) = 1 + 4t,$

$y(t) = 2 - 2t,$

$z(t) = 3 + t.$

Exercise 1

Find the parametric equations of the line through the two points $$(2,1,7)$$ and $$(1,3,5)$$.

Hint: a vector parallel to the line has tail at $$(2,1,7)$$ and head at $$(1,3,5)$$.

### Planes

If S is a plane then a vector n is normal (perpendicular) to the plane if it is orthogonal to every vector that lies on the plane.  Suppose that n is a normal vector to a plane and $$(a,b,c)$$ is a point on the plane.  Let $$(x,y,z)$$ be a general point on the plane, then

$<x - a, y - b, z - c>$

is parallel to the plane, hence

$\vec{n} \cdot <x - a, y - b, z - c> = 0.$

This defines the equation of the plane.

Example 2

Find the equation of the plane that contains the point $$(2,1,0)$$ and has normal vector $$<1,2,3>$$.

Solution

We have

$<1,2,3> \cdot <x - 2,y - 1,z - 0> = 0$

so that

$1(x - 2) + 2(y - 1) + 3z = 0$

or

$x + 2y + 3z = 4.$

Example 3

Find the equation of the plane through the points

• $$P = (0,0,1)$$
• $$Q = (2,1,0)$$
• $$R = (1,1,1)$$

Solution

Let

$\textbf{v} = Q-P = <2,1,-1>$

and

$\textbf{w} = R-P = <1,1,0>$

then to find a vector normal to the plane, we find the cross product of $$v$$ and $$w$$:

$v \times w = \begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{i}} & \hat{\textbf{i}} \\ 2 & 1 &-1 \\ 1 &1 0 \end{vmatrix} = \hat{\textbf{i}} - \hat{\textbf{j}} + \hat{\textbf{k}}$

or

$<1, -1, 1>.$

We can now use the formula:

$<1, -1, 1> \cdot <x, y, z - 1> = 0$

or

$x - y + z - 1 = 0$

or

$x - y + z = 1$

### Distance Between a Point and a Plane

Let P be a point and Q be a point on a plane with normal vector n, then the distance between P and the plane is given by

Note:  Distance Between a Point $$P$$ and a Plane With Normal Vector n

Let $$Q$$ be a point on the plane with normal vector $$\vec{n}$$.  The the distance from the point $$P$$ to this plane is given by

$Proj_nPQ = \dfrac{ ||PQ \cdot \vec{n}|| }{ || \vec{n}|| }.$

Example 4

Find the distance between the point $$(1,2,3)$$ and the plane

$2x - y - 2z = 5.$

Solution

The normal vector can be read off from the equation as

$\vec{n} = <2, -1, -2>.$

Now find a convenient point on the plane such as $$Q = (0, -5, 0)$$.  We have

$Q = <-1, -7, -3>$

and

$\vec{n} \cdot PQ = -2 + 7 + 6 = 11.$

We find the magnitude of n by taking the square root of the sum of the squares.  The sum is

$4 + 1 + 4 = 9$

so

$|| \vec{n} || = 3.$

Hence the distance from the point to the plane is $$\frac{11}{3}$$.

### The Angle Between 2 Planes

The angle between two planes is given by the angle between the normal vectors.

Example

Find the angle between the two planes

$3x - 2y + 5z = 1$

and

$4x + 2y - z = 4.$

We have the two normal vectors are

$\vec{n} = <3,-2,5>$

and

$\vec{m} = <4,2,-1>.$

We have

$\vec{n} \cdot \vec{ m} = 3$

$|| \vec{n} || = \sqrt{38}$

$|| \vec{m} || = \sqrt{21}$

hence the angle is

$\cos^{-1} \left(\dfrac{3}{\sqrt{38}\sqrt{21}} \right) = 1.46 \, rad.$

### Contributors

• Integrated by Justin Marshall.