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Shortest Distance Between Plane and Line
Related Topics
Wize University Linear Algebra Textbook > Equations of Lines and Planes
Distances
5 Activities
Find the shortest distance between the following plane and line in
R
3
\mathbb{R}^3
R
3
:
P
:
2
x
−
3
y
+
z
=
1
,
L
:
{
x
=
2
t
+
3
y
=
t
+
4
z
=
−
t
−
2
P: 2x-3y+z=1, L: \begin{cases} x = 2t+3 \\ y = t + 4 \\ z = -t - 2 \end{cases}
P
:
2
x
−
3
y
+
z
=
1
,
L
:
⎩
⎨
⎧
x
=
2
t
+
3
y
=
t
+
4
z
=
−
t
−
2
14
\sqrt{14}
14
81
14
\frac{81}{14}
14
81
9
14
\frac{9}{\sqrt{14}}
14
9
34
\sqrt{34}
34
I don't know
Check Submission
More Distances Questions:
Find the distance between following planes:
P
1
:
2
x
+
y
+
z
=
5
,
P
2
:
−
6
x
−
3
y
−
3
z
=
25
P1:2x+y+z=5,\ \ \ P2:-6x-3y-3z=25
P
1
:
2
x
+
y
+
z
=
5
,
P
2
:
−
6
x
−
3
y
−
3
z
=
25
Find a point on the plane defined by
3
x
−
y
+
z
=
8
,
3x-y+z=8,
3
x
−
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
∈
R
a,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
Q
⃗
=
(
p
,
q
,
r
)
\vec Q=(p,q,r)
Q
=
(
p
,
q
,
r
)
and the plane
a
x
+
b
y
+
c
z
=
d
ax+by+cz=d
a
x
+
b
y
+
cz
=
d
is
∣
d
−
a
p
−
b
q
−
c
r
∣
a
2
+
b
2
+
c
2
\frac{|d-ap-bq-cr|}{\sqrt{a^2+b^2+c^2}}
a
2
+
b
2
+
c
2
∣
d
−
a
p
−
b
q
−
cr
∣
.
Find the closest distance of point (2,2,2) to the line L by deriving and solving system of linear equations.
L
:
{
𝑥
1
+
x
2
+
𝑥
3
=
2
𝑥
1
−
𝑥
2
−
2
x
3
=
5
L:\begin{cases} 𝑥_1+ x_2 + 𝑥_3 = 2\\ 𝑥_1 − 𝑥_2 − 2x_3 = 5 \end{cases}
L
:
{
x
1
+
x
2
+
x
3
=
2
x
1
−
x
2
−
2
x
3
=
5
Let
L
1
L_1
L
1
be a line through the point
(
1
,
3
,
1
)
(1,3,1)
(
1
,
3
,
1
)
with direction vector
[
1
1
2
]
T
\begin{bmatrix}1&1&2\end{bmatrix}^T
[
1
1
2
]
T
and let
L
2
L_2
L
2
be a line through the point
(
2
,
1
,
2
)
(2,1,2)
(
2
,
1
,
2
)
with direction vector
[
1
1
3
]
T
\begin{bmatrix}1&1&3\end{bmatrix}^T
[
1
1
3
]
T
. Find the shortest distance from
L
1
L_1
L
1
to
L
2
L_2
L
2
.
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
l
1
\bcb{l_1}
l
1
be the line passing through
P
\bcb{P}
P
and
Q
\bcb{Q}
Q
, and let
l
2
\bcb{l_2}
l
2
be the line passing through
R
\bcb{R}
R
and
S
\bcb{S}
S
. Find the distance between
R
\bcb{R}
R
and
l
1
\bcb{l_1}
l
1
. Find the distance between
l
1
\bcb{l_1}
l
1
and
l
2
\bcb{l_2}
l
2
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 60.8
Given the planes
π
1
:
4
x
−
3
y
+
z
=
2
\bcb{\pi_1 \, : \, 4x - 3y + z = 2}
π
1
:
4
x
−
3
y
+
z
=
2
and
π
2
:
x
+
2
y
+
2
z
=
0
\bcb{\pi_2 \, : \, x + 2y + 2z = 0}
π
2
:
x
+
2
y
+
2
z
=
0
, and the lines
L
1
:
{
x
=
3
t
+
2
y
=
−
2
t
+
1
z
=
t
,
L
2
:
{
x
=
−
2
t
+
1
y
=
t
+
1
z
=
t
+
3
,
L
3
:
{
x
=
t
+
3
y
=
−
t
z
=
2
t
+
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.}
L
1
:
⎩
⎨
⎧
x
y
z
=
=
=
3
t
+
2
−
2
t
+
1
t
,
L
2
:
⎩
⎨
⎧
x
y
z
=
=
=
−
2
t
+
1
t
+
1
t
+
3
,
L
3
:
⎩
⎨
⎧
x
y
z
=
=
=
t
+
3
−
t
2
t
+
1
,
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 58.2
Given the vector
v
⃗
=
<
1
,
1
,
2
>
\bcb{\vec{v} = \left< 1, 1, 2 \right>}
v
=
⟨
1
,
1
,
2
⟩
and the points
P
1
=
(
0
,
−
1
,
0
)
\bcb{P_1 = (0,-1,0)}
P
1
=
(
0
,
−
1
,
0
)
and
P
2
=
(
−
1
,
1
,
0
)
\bcb{P_2 = (-1,1,0)}
P
2
=
(
−
1
,
1
,
0
)
, find the
minimum
distance of
P
2
\bcb{P_2}
P
2
to the line
L
\bcb{L}
L
that is parallel to
v
⃗
\bcb{\vec{v}}
v
that passes through
P
1
\bcb{P_1}
P
1
.
Shortest Distance Between Two Lines, Parallel Edition
Find the shortest distance between the following two lines in
R
3
\mathbb{R}^3
R
3
:
L
1
=
{
x
=
1
+
2
t
y
=
3
−
t
z
=
2
+
3
t
,
L
2
=
{
1
−
4
t
2
+
2
t
−
6
t
L_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}
L
1
=
⎩
⎨
⎧
x
=
1
+
2
t
y
=
3
−
t
z
=
2
+
3
t
,
L
2
=
⎩
⎨
⎧
1
−
4
t
2
+
2
t
−
6
t
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
2
x
−
3
y
+
z
=
4
2x-3y+z=4
2
x
−
3
y
+
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
L
⃗
(
t
)
=
<
2
t
+
3
,
t
+
4
,
−
t
−
2
>
\vec{L}(t)=\left<2t+3,t+4,-t-2\right>
L
(
t
)
=
⟨
2
t
+
3
,
t
+
4
,
−
t
−
2
⟩
.
Shortest Distance Between Two Planes
Find the shortest distance between the following two planes in
R
3
\mathbb{R}^3
R
3
:
P
1
:
2
x
−
3
y
+
z
=
1
P
2
:
−
6
x
+
9
y
−
3
z
=
5
P_1: 2x - 3y + z = 1\\ P_2: -6x + 9y - 3z = 5
P
1
:
2
x
−
3
y
+
z
=
1
P
2
:
−
6
x
+
9
y
−
3
z
=
5
Shortest Distance Between Two Lines, Parallel Edition
Find the shortest distance between the following two lines in
R
3
\mathbb{R}^3
R
3
:
L
1
=
{
x
=
1
+
2
t
y
=
3
−
t
z
=
2
+
3
t
,
L
2
=
{
1
−
4
t
2
+
2
t
−
6
t
L_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}
L
1
=
⎩
⎨
⎧
x
=
1
+
2
t
y
=
3
−
t
z
=
2
+
3
t
,
L
2
=
⎩
⎨
⎧
1
−
4
t
2
+
2
t
−
6
t
Shortest Distance Between Two Lines
Find the shortest distance between the following two lines in
R
3
\mathbb{R}^3
R
3
:
L
1
=
{
x
=
1
+
2
t
y
=
3
−
t
z
=
2
+
3
t
,
L
2
=
{
−
t
2
+
t
−
3
t
L_1 = \begin{cases} x = 1+2t \\ y = 3-t \\ z = 2+3t \end{cases}, L_2 = \begin{cases} -t \\ 2+t \\ -3t \end{cases}
L
1
=
⎩
⎨
⎧
x
=
1
+
2
t
y
=
3
−
t
z
=
2
+
3
t
,
L
2
=
⎩
⎨
⎧
−
t
2
+
t
−
3
t
Shortest Distance
Find the shortest distance from the point
(
3
,
−
1
,
2
)
\left(3,\ -1,\ 2\right)
(
3
,
−
1
,
2
)
to the plane
2
x
−
y
+
z
=
10
2x-y+z=10
2
x
−
y
+
z
=
10
.