Wize University Linear Algebra Textbook > Equations of Lines and Planes

Lines in $\reals^n$

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Lines in Rn\colorOne{\reals^n}Rn
Vector Form
A line in Rn\mathbb{R}^nRn can be described by
a point on the line, and
a vector parallel to the line called a direction vector.

Let PPP be a known point on the line, d⃗∈Rn\vec{d} \in \reals^nd∈Rn be a direction vector, and t∈Rt\in\realst∈R be a parameter.
Then every vector x⃗=⟨x1,x2,…,xn⟩\vec x=\lang x_1, x_2, \dots, x_n \rangx=⟨x1​,x2​,…,xn​⟩ on the line can be described in vector form (or point-parallel form) by:
x⃗ = P⃗+td⃗\boxed{\quad
\vec x\ =\ \vec{P}+t\vec{d}
\quad}x = P+td​
Geometrically
Start at point PPP and run along a direction vector d⃗\vec dd to form a line!

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Parametric Form
We can break up the vector form into separate parametric equations.

Example: R2\reals^2R2
Known point on the line: P(x1,y1)P(x_1,y_1)P(x1​,y1​)
Direction vector: d⃗=⟨d1,d2⟩\vec{d}= \lang d_1,d_2 \rangd=⟨d1​,d2​⟩
x⃗=P⃗+td⃗⇓{x=x1+td1y=y1+td2\begin{array}{c}
\vec x = \vec P + t\vec d\\[0.3em]
\Downarrow\\[0.3em]


\begin{cases}
x=x_1+td_1\\
y=y_1+td_2
\end{cases}

\end{array}x=P+td⇓{x=x1​+td1​y=y1​+td2​​​
Symmetric Form
Solving for the parameter for each parametric equation and equating the solutions, we obtain the symmetric form of a line:

x−x1d1=y−y1d2  (=z−z1d3=… )\boxed{\quad
\frac{x-x_1}{d_1} = \frac{y-y_1}{d_2} \ \ \left(=\frac{z-z_1}{d_3} =  \dots\right)
\quad}d1​x−x1​​=d2​y−y1​​  (=d3​z−z1​​=…)​

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Two-Point Form
We can also describe a line using just two points.
Given points PPP and QQQ on the line, we may write the line as:
x⃗=(1−t)P⃗+tQ⃗\boxed{\quad
\vec x = (1-t)\vec P + t \vec Q
\quad}x=(1−t)P+tQ​​

Show that this is equivalent to vector form with d⃗=PQ→\vec d = \overrightarrow{PQ}d=PQ​.

x⃗=(1−t)P⃗+tQ⃗=P⃗−tP⃗+tQ⃗=P⃗+t(Q⃗−P⃗)=P⃗+t PQ→=P⃗+td⃗\begin{aligned}
\vec x 
&= (1-t)\vec P + t \vec Q\\
&= \vec P - t\vec P + t \vec Q\\
&= \vec P + t (\vec Q - \vec P)\\
&= \vec P + t\ \overrightarrow{PQ}\\
&= \vec P + t\vec d
\end{aligned}x​=(1−t)P+tQ​=P−tP+tQ​=P+t(Q​−P)=P+t PQ​=P+td​

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Lines in R2\colorOne{\reals^2}R2
Lines in R2\reals^2R2 have some special forms. Note that these forms describe hyperplanes in Rn\reals^nRn.

Point-Normal Form
Given a point on the line P(x1,y1)P(x_1,y_1)P(x1​,y1​) and a normal (orthogonal) vector n⃗=⟨a,b⟩\vec n = \lang a,b \rangn=⟨a,b⟩:
n⃗⋅(x⃗−P⃗)=0\boxed{\quad
\vec n \cdot (\vec x - \vec P) = 0
\quad}n⋅(x−P)=0​

Wize Concept
Recall: given a direction vector [ab]∈R2\begin{bmatrix}
  a\\ b\\
\end{bmatrix}
 \in \reals^2[ab​]∈R2, we can "switch and flip" to find an orthogonal vector: [b−a]\begin{bmatrix}
  b\\ -a\\
\end{bmatrix}[b−a​]

Standard Form
Given a point on the line P(x1,y1)P(x_1, y_1)P(x1​,y1​)  and a normal vector n⃗=⟨a,b⟩\vec n = \lang a,b\rangn=⟨a,b⟩, and with c=n⃗⋅P⃗c = \vec n \cdot \vec Pc=n⋅P:
ax+by=c\boxed{\quad
ax + by = c
\quad}ax+by=c​
Show that point-normal form is equivalent to standard form.

n⃗⋅(x⃗−P⃗)=0n⃗⋅x⃗−n⃗⋅P⃗=0n⃗⋅x⃗=n⃗⋅P⃗⟨a,b⟩⋅⟨x,y⟩=cax+by=c\begin{aligned}
\vec n \cdot (\vec x - \vec P) &= 0\\
\vec n \cdot \vec x - \vec n \cdot \vec P &= 0\\
\vec n \cdot \vec x &= \vec n \cdot \vec P\\
\lang a,b \rang \cdot \lang x, y \rang &= c\\
ax + by &= c
\end{aligned}n⋅(x−P)n⋅x−n⋅Pn⋅x⟨a,b⟩⋅⟨x,y⟩ax+by​=0=0=n⋅P=c=c​

Wize Tip
Two lines are parallel if their direction vectors are parallel:  d1⃗=kd2⃗ \vec{d_1}=k\vec{d_2}d1​​=kd2​​
Two lines coincide (same line) if they are parallel and they share one point
Two lines are orthogonal if their direction vectors are orthogonal:  d1⃗⋅d2⃗=0\vec{d_1}\cdot\vec{d_2}=0d1​​⋅d2​​=0

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Example: Lines in Rn\colorOne{\reals^n}Rn
Find the vector form, parametric form, and two-point form of the line that passes through the points A(−2,5,8)A\left(-2,5,8\right)A(−2,5,8) and B(−1,−1,3)B\left(-1,-1,3\right)B(−1,−1,3).

Given two points, we can immediately write the two-point form:
x⃗=(1−t)A⃗+tB⃗⟨x,y,z⟩=(1−t)⟨−2,5,8⟩+t⟨−1,−1,3⟩\begin{aligned}
\vec x &= (1-t)\vec A + t\vec B\\[0.5em]
\lang x,y,z \rang &= (1-t)\lang -2,5,8 \rang + t\lang -1,-1,3 \rang\\
\end{aligned}x⟨x,y,z⟩​=(1−t)A+tB=(1−t)⟨−2,5,8⟩+t⟨−1,−1,3⟩​

Using vector form instead, we first need to find the direction vector:
d⃗=AB→ = ⟨−1,−1,3⟩−⟨−2,5,8⟩ = ⟨1,−6,−5⟩\vec{d}=\overrightarrow{AB}
\ =\ \lang -1,-1,3 \rang - \lang-2,5,8\rang
\ =\ \lang 1,-6,-5\rangd=AB = ⟨−1,−1,3⟩−⟨−2,5,8⟩ = ⟨1,−6,−5⟩

Now we can use either point AAA or point BBB as our known point.
Using the known point AAA we have:

x⃗=A⃗+td⃗⟨x,y,z⟩=⟨−2,5,8⟩+t⟨1,−6,−5⟩\begin{aligned}
\vec x &= \vec A + t\vec d\\
\lang \colorOne{x}, \colorTwo{y}, \colorThree{z} \rang &=
\lang \colorOne{-2},\colorTwo{5},\colorThree{8} \rang 
+ 
t\lang\colorOne{1},\colorTwo{-6},\colorThree{-5}\rang
\end{aligned}x⟨x,y,z⟩​=A+td=⟨−2,5,8⟩+t⟨1,−6,−5⟩​

Splitting the vector form into its parametric equations yields:

x=−2+t,y=5−6t,z=8−5t\colorOne{x=-2+t},\quad 
\colorTwo{y=5-6t},\quad 
\colorThree{z=8-5t}x=−2+t,y=5−6t,z=8−5t

Exam Tip
For a multiple choice question, you might have to use BBB instead of AAA if you don't see this answer.

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Example: Lines in R2\colorOne{\reals^2}R2
Find the point-normal and standard form equations of the line that passes through the point P(11,0)P(11,0)P(11,0) and is parallel to the line with parametric equations:
x=−3t,y=2−tx=-3t,\quad y=2-tx=−3t,y=2−t

The given line's parametric equations can be written as:

x=0−3t,y=2−t  ⟹  ⟨x,y⟩=⟨0,2⟩+t⟨−3,−1⟩\begin{aligned}
&
x=\colorOne{0}\colorTwo{-3t},\quad
y=\colorOne{2}\colorTwo{-t}\\[0.5em] 

\implies & \lang x,y\rang 
= \colorOne{\lang 0,2\rang}
+
\colorTwo{t\lang-3,-1\rang}
\end{aligned}⟹​x=0−3t,y=2−t⟨x,y⟩=⟨0,2⟩+t⟨−3,−1⟩​

Since the line we want is parallel to this one, we use the direction vector: d⃗=⟨−3,−1⟩\vec{d}=\colorTwo{\lang-3,-1\rang}
d=⟨−3,−1⟩

We want our line to pass through P(11,0)P(11,0)P(11,0), so the vector form equation of our line is:

⟨x,y⟩=⟨11,0⟩+t⟨−3,−1⟩\lang x,y\rang = \lang 11,0 \rang + t\lang -3,-1\rang⟨x,y⟩=⟨11,0⟩+t⟨−3,−1⟩
Since we want the point-normal and standard form, we need a normal vector:
Use "switch and flip" on d⃗\vec dd to get n⃗=⟨−1,3⟩\vec n = \lang -1,3 \rangn=⟨−1,3⟩.
So the point-normal form is:
n⃗⋅(x⃗−P⃗)=0⟨−1,3⟩⋅(x⃗−⟨11,0⟩)=0\begin{array}{r}
\vec n \cdot (\vec x - \vec P) = 0\\[0.5em]

\boxed{\lang -1, 3 \rang \cdot ( \vec x - \lang 11,0 \rang)
= 0}
\end{array}n⋅(x−P)=0⟨−1,3⟩⋅(x−⟨11,0⟩)=0​​
For the standard form, all that's left to do is calculate c=n⃗⋅P⃗=⟨−1,3⟩⋅⟨11,0⟩=−11c 
= \vec n \cdot \vec P 
= \lang -1,3 \rang \cdot \lang 11,0 \rang
= -11c=n⋅P=⟨−1,3⟩⋅⟨11,0⟩=−11.
So we read off the coefficients from n⃗\vec nn for the LHS, and the RHS is simply c=−11c=-11c=−11:
−x+3y=−11\boxed{-x + 3y = -11}−x+3y=−11​

Practice: Point on Line
Find all values of kkk such that the point P(k+3,k,−5)P(k+3,k,-5)P(k+3,k,−5) lies on the line
l: x⃗=[21−4]+t[131]l: \ \vec x 
=\begin{bmatrix}
  2\\ 1\\ -4\\
\end{bmatrix}
+
t
\begin{bmatrix}
  1\\ 3\\ 1\\
\end{bmatrix}l: x=​21−4​​+t​131​​

k=

I don't know

Practice: Lines in Rn\colorOne{\reals^n}Rn
Various pieces of information can be given to find the equation of a line.

Given Two Points
Find the vector form and standard form of the equation of the line that passes through the points A(−3,0)A\left(-3,0\right)A(−3,0) and B(1,2)B\left(1,2\right)B(1,2).

x⃗=⟨1,2⟩+t⟨2,1⟩\vec{x}=\lang 1,2\rang + t\lang 2,1\rangx=⟨1,2⟩+t⟨2,1⟩;  x−2y=−3x-2y=-3x−2y=−3

x⃗=⟨1,2⟩+t⟨−4,2⟩\vec{x}=\lang 1,2\rang+t\lang-4,2\rangx=⟨1,2⟩+t⟨−4,2⟩;  x+2y=5x+2y=5x+2y=5

x⃗=⟨1,2⟩+t⟨1,−2⟩\vec{x}=\lang 1,2\rang+t\lang1,-2\rangx=⟨1,2⟩+t⟨1,−2⟩;  2x+y=42x+y=42x+y=4

I don't know

Practice: Lines in Rn\colorOne{\reals^n}Rn
Find the parametric equations of the line passing through the point A(3,3,−3)A(3,3,-3)A(3,3,−3) and parallel to the line with equations:
x=1−2t,y=3+t,z=−6tx=1-2t,\quad y=3+t,\quad z=-6tx=1−2t,y=3+t,z=−6t

Answer below with the appropriate coefficients:
x = 
 + 
t
y = 
 + 
t
z = 
 + 
t

I don't know

Extra Practice

Consider u=(1,2,3) and v=(2,-1,4)
a. Find a parametric equation of the line passing through the point (1,3,1) and 
    perpendicular to both u and v.

1) Find a vector of length 1 that is parallel to the line given by the equations:
𝑦−𝑧=42𝑥+𝑦+3𝑧=2\begin{array}{}𝑦−𝑧=4\\2𝑥+𝑦+3𝑧=2

\end{array}y−z=42x+y+3z=2​

a) Write the parametric equation of line 
L:{x+2y=0x−z=5L: \begin{cases}
x+2y=0\\
x-z=5
\end{cases}L:{x+2y=0x−z=5​
b) Find intersection point of the line L with the plane P:3y−3z=2.P:3y-3z=2.P:3y−3z=2.

Find vec form line thru pt ortho to 2 vec form lines

Find the equation of the line that passes through the point (1,2,−3)\left(1,2,-3\right)(1,2,−3) and is orthogonal to both lines
l1: (−5, 0, 1)+t(1,0,−1)l_1:\ \left(-5,\ 0,\ 1\right)+t\left(1,0,-1\right)l1​: (−5, 0, 1)+t(1,0,−1) and l2: (3,−1,6)+t(0,2,1)l_2:\ \left(3,-1,6\right)+t\left(0,2,1\right)l2​: (3,−1,6)+t(0,2,1).

Practice Question: Two Point Form Equation
Write a vector that is parallel to the line (x, y, z)=(1−t)(6,−1,2)+t(9,5,−7)\left(x,\ y,\ z\right)=\left(1-t\right)\left(6,-1,2\right)+t\left(9,5,-7\right)(x, y, z)=(1−t)(6,−1,2)+t(9,5,−7).

Find vec form line thru pt ortho to 2 vec form lines

Find the equation of the line that passes through the point (1,2,−3)\left(1,2,-3\right)(1,2,−3) and is orthogonal to the lines with equations:
 (x,y,z)=(−5,0,1)+t(1,0,−1)(x,y,z)= (-5,0,1)+t(1,0,-1)(x,y,z)=(−5,0,1)+t(1,0,−1) and (x,y,z)=(3,−1,6)+t(0,2,1)(x,y,z)=(3,-1,6)+t(0,2,1)(x,y,z)=(3,−1,6)+t(0,2,1)

Which of the following vectors is a normal vector of a line that is parallel to 2x+y=−32x+y=-32x+y=−3?

133 - FML 3 - 18.1W e.g. 33; 222/262_mid_#2_19.1W_eg5

Find the equations of the line passing through the point (2,3,−4)\bcb{(2, 3, -4)}(2,3,−4) parallel to the vector i−4k\bcb{\mathbf{i} - 4\mathbf{k}}i−4k. Express them in vector-parametric, scalar-parametric and standard form.

133 - FML 3 - 18.1W e.g. 50

 Find the equation of a plane perpendicular to w⃗=<1,−2,1>\bcb{\vec{w} = \left< 1, -2, 1\right>}w=⟨1,−2,1⟩ that contains the point P=(−1,1,2)\bcb{P = (-1, 1, 2)}P=(−1,1,2). 

Lines in $\reals^n$

Practice: Lines in Rn\colorOne{\reals^n}Rn
Various pieces of information can be given to find the equation of a line.

133 - FML 3 - 18.1W e.g. 58.3

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0)
Find the point on the line L\bcb{L}L closest to the point P2\bcb{P_2}P2​.

133 - FML 3 - 18.1W e.g. 58.2

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0)
Find the minimum distance of P2\bcb{P_2}P2​ to the line L\bcb{L}L found above. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 56

Given the points P1=(1,0,5)\bcb{P_1 = (1, 0, 5)}P1​=(1,0,5) and P2=(3,1,−2)\bcb{P_2 = (3, 1, -2)}P2​=(3,1,−2), which of the following points are on the line connecting P1\bcb{P_1}P1​ and P2\bcb{P_2}P2​?

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41

Let P=(−3,−3,−4)\bcb{P = (-3, -3, -4)}P=(−3,−3,−4), Q=(−5,−6,−3)\bcb{Q = (-5,-6,-3)}Q=(−5,−6,−3), R=(−6,−4,−6)\bcb{R = (-6,-4,-6)}R=(−6,−4,−6) and S=(−2,−3,−4)\bcb{S = (-2, -3, -4)}S=(−2,−3,−4). Let l1\bcb{l_1}l1​ be the line passing through P\bcb{P}P and Q\bcb{Q}Q, and let l2\bcb{l_2}l2​ be the line passing through R\bcb{R}R and S\bcb{S}S. Find the distance between R\bcb{R}R and l1\bcb{l_1}l1​. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.1

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0), find the scalar-parametric equation of the line L\bcb{L}L  parallel to v⃗\bcb{\vec{v}}v that passes through P1\bcb{P_1}P1​.

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2

Given the vector v⃗=<1,1,2>\bcb{\vec{v} = \left< 1, 1, 2 \right>}v=⟨1,1,2⟩ and the points P1=(0,−1,0)\bcb{P_1 = (0,-1,0)}P1​=(0,−1,0) and P2=(−1,1,0)\bcb{P_2 = (-1,1,0)}P2​=(−1,1,0), find the minimum distance of P2\bcb{P_2}P2​ to the line L\bcb{L}L that is parallel to v⃗\bcb{\vec{v}}v that passes through P1\bcb{P_1}P1​. 

133 - FML 3 - 18.1W - e.g. 8

If a line is given by y=mx\bcb{\boldsymbol{ y = mx}}y=mx, find a unit vector pointing in the direction of the line (i.e. find the directional vector).

Equation of Lines

Which of of the following is an equation of the line that passes through A(0,−3)A(0,-3)A(0,−3) and B(−3,6)B(-3,6)B(−3,6).

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Lines in $\reals^n$

Popular Courses

Lines in $\colorOne{\reals^n}$

Vector Form

Parametric Form

Two-Point Form

Lines in $\colorOne{\reals^2}$

Example: Lines in $\colorOne{\reals^n}$

Example: Lines in $\colorOne{\reals^2}$

Practice: Point on Line

Practice: Lines in $\colorOne{\reals^n}$

Given Two Points

Practice: Lines in $\colorOne{\reals^n}$

Extra Practice

Find vec form line thru pt ortho to 2 vec form lines

Find vec form line thru pt ortho to 2 vec form lines

133 - FML 3 - 18.1W e.g. 33; 222/262_mid_#2_19.1W_eg5

133 - FML 3 - 18.1W e.g. 50

Lines in $\reals^n$

133 - FML 3 - 18.1W e.g. 58.3

133 - FML 3 - 18.1W e.g. 58.2

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 56

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.1

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2

133 - FML 3 - 18.1W - e.g. 8

Equation of Lines

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Lines in Rn\reals^nRn

Popular Courses

Lines in Rn\colorOne{\reals^n}Rn

Vector Form

Parametric Form

Two-Point Form

Lines in R2\colorOne{\reals^2}R2

Example: Lines in Rn\colorOne{\reals^n}Rn

Example: Lines in R2\colorOne{\reals^2}R2

Practice: Point on Line

Practice: Lines in Rn\colorOne{\reals^n}Rn

Given Two Points

Practice: Lines in Rn\colorOne{\reals^n}Rn

Extra Practice

Find vec form line thru pt ortho to 2 vec form lines

Find vec form line thru pt ortho to 2 vec form lines

133 - FML 3 - 18.1W e.g. 33; 222/262_mid_#2_19.1W_eg5

133 - FML 3 - 18.1W e.g. 50

Lines in $\reals^n$

133 - FML 3 - 18.1W e.g. 58.3

133 - FML 3 - 18.1W e.g. 58.2

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 56

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 41

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.1

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2

133 - FML 3 - 18.1W - e.g. 8

Equation of Lines

Lines in $\reals^n$

Lines in $\colorOne{\reals^n}$

Lines in $\colorOne{\reals^2}$

Example: Lines in $\colorOne{\reals^n}$

Example: Lines in $\colorOne{\reals^2}$

Practice: Lines in $\colorOne{\reals^n}$

Practice: Lines in $\colorOne{\reals^n}$