1.6: Lines and Planes

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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

\[ \langle x,y \rangle = P + tv\nonumber \]

for some number $t$.

$linpln20.gif$

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

Parametric Equations of a Line

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

\[\langle x,y,z\rangle = \langle a,b,c\rangle + t\textbf{v}.\nonumber \]

Example $\PageIndex{1}$

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

Solution

We write:

\[\langle x, y, z\rangle = \langle 1, 2, 3\rangle + t \langle 4, -2, 1\rangle = \langle 1 + 4t, 2 - 2t, 3 + t\rangle \nonumber \]

\[ x(t) = 1 + 4t,\nonumber \]

\[y(t) = 2 - 2t,\nonumber \]

\[z(t) = 3 + t.\nonumber \]

Exercise $\PageIndex{1}$

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

\[ \langle x - a, y - b, z - c\rangle \nonumber \]

is parallel to the plane, hence

\[\vec{n} \cdot \langle x - a, y - b, z - c\rangle = 0.\nonumber \]

This defines the equation of the plane.

$linpln21.gif$

Example $\PageIndex{2}$

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

Solution

We have

\[\langle 1,2,3\rangle \cdot \langle x - 2,y - 1,z - 0\rangle = 0\nonumber \]

so that

\[1(x - 2) + 2(y - 1) + 3z = 0\nonumber \]

\[ x + 2y + 3z = 4.\nonumber \]

Example $\PageIndex{3}$

Find the equation of the plane through the points

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

Solution

Let

\[\textbf{v} = Q-P = \langle 2,1,-1\rangle \nonumber \]

and

\[\textbf{w} = R-P = \langle 1,1,0\rangle \nonumber \]

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

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

\[\langle 1, -1, 1\rangle .\nonumber \]

We can now use the formula:

\[\langle 1, -1, 1\rangle \cdot \langle x, y, z - 1\rangle = 0\nonumber \]

\[x - y + z - 1 = 0\nonumber \]

\[x - y + z = 1\nonumber \]

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

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

\[ Proj_nPQ = \dfrac{ ||PQ \cdot \vec{n}|| }{ || \vec{n}|| }.\nonumber \]

Example $\PageIndex{4}$

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

\[2x - y - 2z = 5.\nonumber \]

Solution

The normal vector can be read off from the equation as

\[\vec{n} = \langle 2, -1, -2\rangle .\nonumber \]

$linpln25.gif$

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

\[Q = \langle -1, -7, -3\rangle \nonumber \]

and

\[ \vec{n} \cdot PQ = -2 + 7 + 6 = 11.\nonumber \]

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

\[4 + 1 + 4 = 9\nonumber \]

\[|| \vec{n} || = 3.\nonumber \]

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

The Angle Between Two Planes

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

Example $\PageIndex{5}$

Find the angle between the two planes

\[ 3x - 2y + 5z = 1\nonumber \]

and

\[4x + 2y - z = 4.\nonumber \]

We have the two normal vectors are

\[ \vec{n} = \langle 3,-2,5\rangle \nonumber \]

and

\[\vec{m} = \langle 4,2,-1\rangle .\nonumber \]

We have

\[\vec{n} \cdot \vec{ m} = 3\nonumber \]

\[ || \vec{n} || = \sqrt{38}\nonumber \]

\[ || \vec{m} || = \sqrt{21}\nonumber \]

hence the angle is

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

Contributors and Attributions

Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.