Wize University Linear Algebra Textbook > Equations of Lines and Planes

Distances

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Distances
Distance Between a Point and a Line
To calculate the shortest distance between a point and a line we use the perpendicular perp\text{perp}perp(based on the projection proj\text{proj}proj).

Given a point Q(q1,q2,…,qn)Q(q_1, q_2, \dots, q_n)Q(q1​,q2​,…,qn​) and a line in vector form  l: x⃗=P⃗+t d⃗l: \ \vec x = \vec P + t\ \vec dl: x=P+t d, the distance from point QQQ to line lll is given by:
distance(Q,l)=∥perpd⃗ PQ→∥\boxed{\quad
\text{distance}(Q,l) = \left\lVert \text{perp}_{\displaystyle \vec d}\ \overrightarrow{PQ} \right\rVert
\quad}distance(Q,l)=​perpd​ PQ​​​
Steps
Find the vector PQ→=Q⃗−P⃗\overrightarrow{PQ} = \vec Q - \vec PPQ​=Q​−P
Find the projection of PQ→\overrightarrow{PQ}PQ​ onto the line's direction vector d⃗\vec{d}d:
projd⃗ PQ→ = (PQ→⋅d⃗d⃗⋅d⃗)d⃗\text{proj}_{\displaystyle \vec d}\  \overrightarrow{PQ} \ =\ 
\left(\dfrac{\overrightarrow{PQ} \cdot \vec d}{\vec d \cdot \vec d}\right) 
\vec dprojd​ PQ​ = (d⋅dPQ​⋅d​)d
Find the perpendicular vector:
perpd⃗ PQ→ = PQ→−projd⃗ PQ→\text{perp}_{\displaystyle \vec d} \ \overrightarrow{PQ} \ =\ \overrightarrow{PQ}-\text{proj}_{\displaystyle \vec{d}}\ \overrightarrow{PQ}perpd​ PQ​ = PQ​−projd​ PQ​
The shortest distance from point QQQ to line lll is the norm of the perpendicular:
distance(Q,l)=∥perpd⃗PQ→∥\text{distance}(Q,l) = \left\lVert \text{perp}_{\displaystyle \vec d} \overrightarrow{PQ} \right\rVertdistance(Q,l)=​perpd​PQ​​
Wize Tip
These same steps can be used to find the shortest distance between a line and a second parallel line:
→\rightarrow→ Choose any point on the second line and call it QQQ.

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Geometrically



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Distance Between a Point and a Hyperplane
We use a slightly different method to find the shortest distance between a point and a hyperplane in Rn\reals^nRn.
Given a point Q(q1,q2,…,qn)Q(q_1, q_2, \dots, q_n)Q(q1​,q2​,…,qn​) and a plane in standard form  Π: a1x1+a2x2+⋯+anxn=b\Pi: \ a_1 x_1 + a_2 x_2 + \dots + a_n x_n = bΠ: a1​x1​+a2​x2​+⋯+an​xn​=b, the distance from point QQQ to plane Π\PiΠ is given by:

distance(Q,Π)=∥projn⃗ PQ→∥\boxed{\quad
\text{distance}(Q,\Pi)
=
\lVert
\text{proj}_{\displaystyle \vec{n}}\ \overrightarrow{PQ}
\rVert
\quad}distance(Q,Π)=∥projn​ PQ​∥​

Steps
Find any point PPP on the plane (it's often easiest to let all but one component be 0 and solve for the third).
Find the vector PQ→=Q⃗−P⃗\overrightarrow{PQ} = \vec Q - \vec PPQ​=Q​−P
Find the projection of PQ→\overrightarrow{PQ}PQ​ onto the hyperplane's normal vector n⃗=⟨a1,a2,…,an⟩\vec{n} = \lang a_1, a_2, \dots, a_n\rangn=⟨a1​,a2​,…,an​⟩ :
projn⃗ PQ→ = (PQ→⋅n⃗n⃗⋅n⃗)n⃗\text{proj}_{\displaystyle \vec n}\  \overrightarrow{PQ} \ =\ 
\left(\dfrac{\overrightarrow{PQ} \cdot \vec n}{\vec n \cdot \vec n}\right) 
\vec nprojn​ PQ​ = (n⋅nPQ​⋅n​)n
The shortest distance from point QQQ to hyperplane Π\PiΠ is the norm of the projection:
distance(Q,Π)=∥projn⃗ PQ→∥\text{distance}(Q, \Pi) = \left\lVert \text{proj}_{\displaystyle \vec n}\  \overrightarrow{PQ} \right\rVertdistance(Q,Π)=​projn​ PQ​​
Wize Tip
These same steps can be used to find the distance between a plane and:
a parallel line, or
a second parallel plane
→\rightarrow→ Choose any point on the line or second plane and call it QQQ.

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Geometrically

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Example: Distance Between Parallel Lines
Find the shortest distance between the parallel lines:
l1: ⟨x,y,z⟩=⟨4,11,5⟩+t⟨1,1,−2⟩l_1:\ \lang x,y,z\rang=\lang4,11,5\rang+t\lang1,1,-2\rangl1​: ⟨x,y,z⟩=⟨4,11,5⟩+t⟨1,1,−2⟩

l2: ⟨x,y,z⟩=⟨0,9,8⟩+s⟨−1,−1,2⟩l_2:\ \lang x,y,z \rang = \lang 0,9,8\rang + s\lang -1,-1,2 \rangl2​: ⟨x,y,z⟩=⟨0,9,8⟩+s⟨−1,−1,2⟩ 

Watch Out!
We may use the method of finding the distance between a point and a line.
This only works for finding the distance between two lines when they are parallel/collinear. 

Distance Between Parallel Lines
Note that the direction vector d⃗2=−d⃗1\vec d_2 = -\vec d_1d2​=−d1​, so the lines are indeed collinear. ✓\colorThree{\checkmark}✓
Since the lines are parallel, we can use any point on l2l_2l2​: choose Q(0,9,8)Q(0,9,8)Q(0,9,8).
Then the problem boils down to finding the distance between point QQQ and line l1l_1l1​. (We can forget about l2l_2l2​ entirely!)

Distance Between a Point and a Line
Choose any point on l1l_1l1​: P(4,11,5)P(4,11,5)P(4,11,5)
PQ→=⟨0,9,8⟩−⟨4,11,5⟩=⟨−4,−2,3⟩\overrightarrow{PQ}=\lang 0,9,8\rang-\lang 4,11,5\rang = \lang -4,-2,3\rang PQ​=⟨0,9,8⟩−⟨4,11,5⟩=⟨−4,−2,3⟩
Find the projection of PQ→\overrightarrow{PQ} 
PQ​ onto the direction vector of the line  d⃗=⟨1,1,−2⟩\vec{d}=\lang 1,1,-2 \rangd=⟨1,1,−2⟩:
projd⃗ PQ→=(PQ→⋅d⃗d⃗⋅d⃗)d⃗=(⟨−4,−2,3⟩⋅⟨1,1,−2⟩⟨1,1,−2⟩⋅⟨1,1,−2⟩)⟨1,1,−2⟩=−126⟨1,1,−2⟩=⟨−2,−2,4⟩\begin{aligned}
\text{proj}_{\displaystyle \vec{d}}\ \overrightarrow{PQ} 

&=
\left(
\dfrac{\overrightarrow{PQ}\cdot\vec d}{\vec d \cdot \vec d}
\right)
\vec d\\[1em]

&=
\left(
\dfrac{\lang-4,-2,3\rang\cdot\lang1,1,-2\rang}{\lang1,1,-2\rang \cdot \lang 1,1,-2\rang}
\right)
\lang 1,1,-2\rang\\[1em]

&=\dfrac{-12}{6} \lang 1,1,-2 \rang\\[1em]

&= \lang -2,-2,4 \rang\\[1em]

\end{aligned}projd​ PQ​​=(d⋅dPQ​⋅d​)d=(⟨1,1,−2⟩⋅⟨1,1,−2⟩⟨−4,−2,3⟩⋅⟨1,1,−2⟩​)⟨1,1,−2⟩=6−12​⟨1,1,−2⟩=⟨−2,−2,4⟩​
Subtract the projection from PQ→\overrightarrow{PQ}PQ​ to find the perpendicular vector:
perpd⃗ PQ→  =  PQ→−projd⃗ PQ→=  ⟨−4,−2,3⟩−⟨−2,−2,4⟩=  ⟨−2,0,−1⟩\begin{aligned}
\text{perp}_{\displaystyle \vec{d}}\ \overrightarrow{PQ} 

\ \ &=\ \  
\overrightarrow{PQ} 
- \text{proj}_{\displaystyle \vec{d}}\ \overrightarrow{PQ}\\[0.5em]

&=\ \  
\lang -4,-2,3\rang 
-\lang -2,-2,4 \rang\\[0.5em]

&=\ \  
\lang -2,0,-1\rang 
\end{aligned}perpd​ PQ​  ​=  PQ​−projd​ PQ​=  ⟨−4,−2,3⟩−⟨−2,−2,4⟩=  ⟨−2,0,−1⟩​


The norm of this vector is the shortest distance between the parallel lines:
∥perpd⃗ PQ→∥=∥⟨−2,0,−1⟩∥=(−2)2+02+(−1)2=5\begin{aligned}
\left\lVert
\text{perp}_{\displaystyle \vec{d}}\ \overrightarrow{PQ} 
\right\rVert
&=  \lVert \lang -2,0,-1\rang \rVert\\
&=  \sqrt{(-2)^2 + 0^2 +(-1)^2}\\
&= \boxed{\sqrt5}
\end{aligned}​perpd​ PQ​​​=∥⟨−2,0,−1⟩∥=(−2)2+02+(−1)2​=5​​​

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Example: Distance Between a Point and a Plane
Find the shortest distance between the plane Π: x−3y−z=−6\Pi: \ x-3y-z=-6Π: x−3y−z=−6 and the point  A(1,1,2)A(1,1,2)A(1,1,2).

The normal vector to the plane can be read from the LHS: n⃗=⟨1,−3,−1⟩\vec{n}= \lang 1,-3,-1 \rangn=⟨1,−3,−1⟩

1. Choose any point on the plane:
Setting y=z=0y=z=0y=z=0 we can easily solve to find x=−6x=-6x=−6 . This gives us the point P(−6,0,0)P(-6,0,0)P(−6,0,0).
2. Find PA→=⟨1,1,2⟩−⟨−6,0,0⟩=⟨7,1,2⟩\overrightarrow{PA}
=\lang 1,1,2\rang - \lang -6,0,0\rang 
=\lang 7,1,2 \rangPA=⟨1,1,2⟩−⟨−6,0,0⟩=⟨7,1,2⟩
3. Project PA→\overrightarrow{PA}PA onto n⃗\vec{n}n:

projn⃗ PA→=(PA→⋅n⃗n⃗⋅n⃗)n⃗=(⟨7,1,2⟩⋅⟨1,−3,−1⟩⟨1,−3,−1⟩⋅⟨1,−3,−1⟩)⟨1,−3,−1⟩=211⟨1,−3,−1⟩=⟨211, −611, −211⟩\begin{aligned}
\text{proj}_{\displaystyle \vec{n}}\ \overrightarrow{PA}
&=
\left(
\dfrac{\overrightarrow{PA}\cdot\vec n}{\vec n \cdot \vec n}
\right)
\vec n\\[1em]

&=
\left(
\dfrac{\lang 7,1,2\rang\cdot\lang 1,-3,-1\rang}{\lang 1,-3,-1\rang \cdot \lang 1,-3,-1\rang}
\right)\lang 1,-3,-1\rang\\[1em]

&=\frac{2}{11}\lang 1,-3,-1\rang\\[1em]

&= \left\lang \dfrac{2}{11},\ -\dfrac{6}{11},\ -\dfrac{2}{11}\right\rang\\[1em]

\end{aligned}projn​ PA​=(n⋅nPA⋅n​)n=(⟨1,−3,−1⟩⋅⟨1,−3,−1⟩⟨7,1,2⟩⋅⟨1,−3,−1⟩​)⟨1,−3,−1⟩=112​⟨1,−3,−1⟩=⟨112​, −116​, −112​⟩​

4. The norm of this vector is the distance between AAA and the plane:

distance(A,Π)=∥⟨211,−611,−211⟩∥=(211)2+(−611)2+(−211)2=4+36+4121=4411=21111\begin{array}{rcl}
\text{distance}(A, \Pi)&=&\left\lVert \left\lang \dfrac{2}{11},-\dfrac{6}{11},-\dfrac{2}{11}\right\rang \right\rVert\\[1em]
&=& \sqrt{\left(\dfrac{2}{11}\right)^2 + \left(-\dfrac{6}{11}\right)^2 + \left(-\dfrac{2}{11}\right)^2}\\[1em]
&=& \sqrt{\dfrac{4+36+4}{121}}\\[1em]
&=&\dfrac{\sqrt{44}}{11}\\[1em]
&=&\dfrac{2\sqrt{11}}{11}\\[1em]
\end{array}distance(A,Π)​=====​​⟨112​,−116​,−112​⟩​(112​)2+(−116​)2+(−112​)2​1214+36+4​​1144​​11211​​​

Practice: Distance Between a Point and a Line
Find the distance between the point P(12,7,−8)P(12,7,-8)P(12,7,−8) and the line that passes through the points A(13,7,−13)A(13,7,-13)A(13,7,−13) and B(19,7,−10)B(19,7,-10)B(19,7,−10).

Distance =

I don't know

Practice: Distance Between Parallel Planes
Find the distance between the two parallel planes  5x−y−2z=65x-y-2z=65x−y−2z=6  and   5x−y−2z=−45x-y-2z=-45x−y−2z=−4.

Distance =

I don't know

Extra Practice

Find the distance between following planes:
P1:2x+y+z=5,   P2:−6x−3y−3z=25P1:2x+y+z=5,\ \ \ P2:-6x-3y-3z=25P1:2x+y+z=5,   P2:−6x−3y−3z=25

Find a point on the plane defined by 3x−y+z=8,3x-y+z=8,3x−y+z=8, which is closest to (2,0,5)

Prove formula for distance between point and plane

Let a,b,c,d,p,q,r∈Ra,b,c,d,p,q,r\in\mathbb{R}a,b,c,d,p,q,r∈R be constants.  Prove that the formula for the distance between the point Q⃗=(p,q,r)\vec Q=(p,q,r)Q​=(p,q,r) and the plane ax+by+cz=dax+by+cz=dax+by+cz=d is ∣d−ap−bq−cr∣a2+b2+c2\frac{|d-ap-bq-cr|}{\sqrt{a^2+b^2+c^2}}a2+b2+c2​∣d−ap−bq−cr∣​ .

Find the closest distance of point (2,2,2) to the line L by deriving and solving system of linear equations.
L:{𝑥1+x2+𝑥3=2𝑥1−𝑥2−2x3=5L:\begin{cases}
𝑥_1+ x_2 + 𝑥_3 = 2\\
𝑥_1 − 𝑥_2 − 2x_3 = 5

\end{cases}L:{x1​+x2​+x3​=2x1​−x2​−2x3​=5​

Let L1L_1L1​ be a line through the point (1,3,1)(1,3,1)(1,3,1) with direction vector
[112]T\begin{bmatrix}1&1&2\end{bmatrix}^T[1​1​2​]Tand let L2L_2L2​ be a line through the point (2,1,2)(2,1,2)(2,1,2) with direction vector [113]T\begin{bmatrix}1&1&3\end{bmatrix}^T[1​1​3​]T . Find the shortest distance from L1L_1L1​ to L2L_2L2​.

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​. Find the distance between l1\bcb{l_1}l1​ and l2\bcb{l_2}l2​. 

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.8

 Given the planes
π1 : 4x−3y+z=2\bcb{\pi_1 \, : \, 4x - 3y + z = 2}π1​:4x−3y+z=2 and π2 : x+2y+2z=0\bcb{\pi_2 \, : \, x + 2y + 2z = 0}π2​:x+2y+2z=0, and the lines
L1 : {x=3t+2y=−2t+1z=t, L2 : {x=−2t+1y=t+1z=t+3, L3 : {x=t+3y=−tz=2t+1\bcb{L_1 \, : \, \left\{ \begin{matrix} x & = & 3t + 2 \\ y & = & -2t + 1 \\ z & = & t \end{matrix} \right., ~ 
L_2 \, : \, \left\{ \begin{matrix} x & = & -2t + 1 \\ y & = & t + 1 \\ z & = & t + 3 \end{matrix} \right., ~ 
L_3 \, : \, \left\{ \begin{matrix} x & = & t + 3 \\ y & = & -t \\ z & = & 2t +1 \end{matrix} \right.}L1​:⎩⎨⎧​xyz​===​3t+2−2t+1t​, L2​:⎩⎨⎧​xyz​===​−2t+1t+1t+3​, L3​:⎩⎨⎧​xyz​===​t+3−t2t+1​,

$\tkct{cut from 19.4F}$ 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​. 

Shortest Distance Between Two Lines, Parallel Edition

Find the shortest distance between the following two lines in R3\mathbb{R}^3R3:
L1={x=1+2ty=3−tz=2+3t,L2={1−4t2+2t−6tL_1 = \begin{cases}
x = 1+2t \\
y = 3-t \\
z = 2+3t
\end{cases}, L_2 = \begin{cases}
1-4t \\
2+2t \\
-6t
\end{cases}L1​=⎩⎨⎧​x=1+2ty=3−tz=2+3t​,L2​=⎩⎨⎧​1−4t2+2t−6t​

Shortest Distance Between Point and Plane

Find the shortest distance between the point (1,2,3)(1,2,3)(1,2,3) and the plane 2x−3y+z=42x-3y+z=42x−3y+z=4.

Shortest Distance Between Point and Line

Find the shortest distance between the point (1,2,3)(1,2,3)(1,2,3) and the line L⃗(t)=<2t+3,t+4,−t−2>\vec{L}(t)=\left<2t+3,t+4,-t-2\right>L(t)=⟨2t+3,t+4,−t−2⟩.

Shortest Distance Between Two Planes

Find the shortest distance between the following two planes in R3\mathbb{R}^3R3:
P1:2x−3y+z=1P2:−6x+9y−3z=5P_1: 2x - 3y + z = 1\\
P_2: -6x + 9y - 3z = 5P1​:2x−3y+z=1P2​:−6x+9y−3z=5

Shortest Distance Between Two Lines, Parallel Edition

Find the shortest distance between the following two lines in R3\mathbb{R}^3R3:
L1={x=1+2ty=3−tz=2+3t,L2={1−4t2+2t−6tL_1 = \begin{cases}
x = 1+2t \\
y = 3-t \\
z = 2+3t
\end{cases}, L_2 = \begin{cases}
1-4t \\
2+2t \\
-6t
\end{cases}L1​=⎩⎨⎧​x=1+2ty=3−tz=2+3t​,L2​=⎩⎨⎧​1−4t2+2t−6t​

Shortest Distance Between Two Lines

Find the shortest distance between the following two lines in R3\mathbb{R}^3R3:
L1={x=1+2ty=3−tz=2+3t,L2={−t2+t−3tL_1 = \begin{cases}
x = 1+2t \\
y = 3-t \\
z = 2+3t
\end{cases}, L_2 = \begin{cases}
-t \\
2+t \\
-3t
\end{cases}L1​=⎩⎨⎧​x=1+2ty=3−tz=2+3t​,L2​=⎩⎨⎧​−t2+t−3t​

Shortest Distance Between Plane and Line

Find the shortest distance between the following plane and line in R3\mathbb{R}^3R3:
P:2x−3y+z=1,L:{x=2t+3y=t+4z=−t−2P: 2x-3y+z=1, L: \begin{cases}
x = 2t+3 \\
y = t + 4 \\
z = -t - 2
\end{cases}P:2x−3y+z=1,L:⎩⎨⎧​x=2t+3y=t+4z=−t−2​

Shortest Distance

Find the shortest distance from the point (3, −1, 2)\left(3,\ -1,\ 2\right)(3, −1, 2) to the plane 2x−y+z=102x-y+z=102x−y+z=10.

Wize University Linear Algebra Textbook > Equations of Lines and Planes

Distances

Popular Courses

Distances

Distance Between a Point and a Line

Distance Between a Point and a Hyperplane

Example: Distance Between Parallel Lines

Distance Between Parallel Lines

Distance Between a Point and a Line

Example: Distance Between a Point and a Plane

Practice: Distance Between a Point and a Line

Practice: Distance Between Parallel Planes

Extra Practice

Prove formula for distance between point and plane

133 - FML 3 - 18.1W e.g. 41

$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.8

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

Shortest Distance Between Two Lines, Parallel Edition

Shortest Distance Between Point and Plane

Shortest Distance Between Point and Line

Shortest Distance Between Two Planes

Shortest Distance Between Two Lines, Parallel Edition

Shortest Distance Between Two Lines

Shortest Distance Between Plane and Line

Shortest Distance