Chapter 10 Part 3
3. Distances
3.1 Distance from point to line:
There are two ways to find the distance from point Q
to the line passing through points P and R
a) Using dot product:
⃗⃗⃗⃗⃗⃗
𝑷𝑸
𝑷𝑹
⃗⃗⃗⃗⃗⃗ ||𝟐 − (𝒄𝒐𝒎𝒑 )𝟐
𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = √||𝑷𝑸
⃗⃗⃗⃗⃗⃗
b) Using cross product:
𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 =
⃗⃗⃗⃗⃗⃗ × 𝑷𝑹
⃗⃗⃗⃗⃗⃗ ||
||𝑷𝑸
⃗⃗⃗⃗⃗⃗ ||
||𝑷𝑹
__________________________________________________________________
Example:
Find the distance from 𝑸(𝟏, 𝟐, 𝟏) to the line through 𝑷(𝟏, 𝟏, 𝟏) and 𝑹(𝟏, 𝟎, −𝟏):
a. using dot product?
b. using cross product?
Solution:
a. Using dot product:
⃗⃗⃗⃗⃗ =< 0,1,0 >
𝑃𝑄
⃗⃗⃗⃗⃗ =< 0, −1, −2 >
𝑃𝑅
1
⃗⃗⃗⃗⃗ ‖ = √1 = 1
‖𝑃𝑄
⃗⃗⃗⃗⃗
𝑃𝑄
𝑐𝑜𝑚𝑝⃗⃗⃗⃗⃗
=
𝑃𝑅
⃗⃗⃗⃗⃗ . 𝑃𝑅
⃗⃗⃗⃗⃗ |
1
|𝑃𝑄
=
⃗⃗⃗⃗⃗ ‖
‖𝑃𝑅
√5
2
4
2
⃗⃗⃗⃗⃗
𝑃𝑄
2
√
⃗⃗⃗⃗⃗
√
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑄 𝑡𝑜 𝑡ℎ𝑒 𝑙𝑖𝑛𝑒 = ||𝑃𝑄|| − (𝑐𝑜𝑚𝑝𝑃𝑅
=
=
)
⃗⃗⃗⃗⃗
5
√5
__________________________________________________________________
b. Using cross product:
⃗⃗⃗⃗⃗ = 𝑄 − 𝑃 =< 0,1,0 >
𝑃𝑄
⃗⃗⃗⃗⃗ = 𝑅 − 𝑃 =< 0, −1, −2 >
𝑃𝑅
_______________________________
𝑖
𝑗
𝑘
⃗⃗⃗⃗⃗ × 𝑃𝑅
⃗⃗⃗⃗⃗ = |0 1
𝑃𝑄
0 | = −2𝑖 = < −2,0,0 >
0 −1 −2
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
⃗⃗⃗⃗⃗ × 𝑃𝑅
⃗⃗⃗⃗⃗ ||
||𝑃𝑄
⃗⃗⃗⃗⃗ ||
||𝑃𝑅
=
2
√5
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Example:
Find the distance from a point 𝑷(𝟐, −𝟔, 𝟏) to the line whose parametric equations
are
𝒍: 𝒙 = 𝟑 + 𝟒𝒕,
𝒚 = 𝟒 − 𝟓𝒕,
a. using dot product?
b. using cross product?
2
𝒛 = −𝟐 + 𝟕𝒕.
Solution:
a. Using dot product:
From line equations, one can find a point 𝑄(3,4, −2) and a vector 𝑎 =< 4, −5,7 >
_______________________________
⃗⃗⃗⃗⃗ = 𝑄 − 𝑃 =< 1,10, −3 >
𝑃𝑄
2
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = √||𝑃𝑄|| − (
⃗⃗⃗⃗⃗ . 𝑎|
67 2
|𝑃𝑄
2
) = √110 − (
) = 7.75
‖𝑎‖
√90
__________________________________________________________________
b. Using cross product:
From line equations, one can find a point 𝑄(3,4, −2) and a vector 𝑎 =< 4, −5,7 >
_______________________________
⃗⃗⃗⃗⃗ = 𝑄 − 𝑃 =< 1,10, −3 >
𝑃𝑄
𝑖
⃗⃗⃗⃗⃗ × 𝑎 = |1
𝑃𝑄
4
𝑗
𝑘
10 −3| = 55𝑖 − 19𝑗 − 45𝑘 = < 55, −19, −45 >
−5 7
||𝑎|| = √16 + 25 + 49 = √90
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
⃗⃗⃗⃗⃗ × 𝑎||
||𝑃𝑄
||𝑎||
=
√5411
√90
= 7.75
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3
3.2 Distance from point to plane:
To find the distance from point 𝑷(𝒙𝟏 , 𝒚𝟏 , 𝒛𝟏 ) to the plane given by
𝑷𝟏 : 𝒂𝟏 𝒙 + 𝒂𝟐 𝒚 + 𝒂𝟑 𝒛 + 𝒄𝟏 = 𝟎:
𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 =
|𝒂𝟏 𝒙𝟏 + 𝒂𝟐 𝒚𝟏 + 𝒂𝟑 𝒛𝟏 + 𝒄𝟏 | |𝒂𝟏 𝒙𝟏 + 𝒂𝟐 𝒚𝟏 + 𝒂𝟑 𝒛𝟏 + 𝒄𝟏 |
=
||𝒂||
√𝒂𝟐𝟏 + 𝒂𝟐𝟐 + 𝒂𝟐𝟑
__________________________________________________________________
Example:
Find the distance from 𝑷(𝟐, −𝟓, 𝟏) to the plane given by equation 𝟐𝒙 − 𝟑𝒚 + 𝒛 =
𝟗?
Solution:
From the plane equation: 𝑎 =< 2, −3,1 >
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
|2(2) − 3(−5) + (1) − 9|
√22 + 32 + 12
=
11
√14
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4
3.3 Distance between two parallel planes:
To find the distance between two parallel planes:
1. Find a point 𝒑𝟏 on the first plane and 𝒑𝟐 on the second plane by
setting the values of two axes to zero (let’s say 𝒚 = 𝒛 = 𝟎).
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
2. Find a vector 𝒑𝟏𝒑𝟐
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝒑 𝒑𝟐
𝟏
3. 𝑫𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝒄𝒐𝒎𝒑⃗⃗⃗⃗⃗
𝒏𝟏
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝒑 𝒑
𝟏 𝟐
= 𝒄𝒐𝒎𝒑⃗⃗⃗⃗⃗
𝒏𝟐 ,
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝒑 𝒑𝟐
𝟏
remember that, 𝒄𝒐𝒎𝒑⃗⃗⃗⃗⃗
𝒏𝟏
=
|𝒑
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗𝟏 |
𝟏 𝒑𝟐 .𝒏
‖𝒏
⃗⃗⃗⃗⃗𝟏 ‖
__________________________________________________________________
Example:
Show that 𝑷𝟏 : −𝟔𝒙 + 𝟑𝒚 − 𝟗𝒛 = 𝟒 and 𝑷𝟐 : 𝟒𝒙 − 𝟐𝒚 + 𝟔𝒛 = 𝟑
are two parallel planes and find the distance between them?
Solution:
𝑛1 =< −6,3, −9 > , 𝑛2 =< 4, −2,6 >
−6
3
−9
=
=
=> 𝑃1 //𝑃2
4
−2
6
_______________________________
Let 𝑦 = 𝑧 = 0 => from 𝑝1 : −6𝑥 = 4 => 𝑥 =
From 𝑃2 : 4𝑥 = 3 => 𝑥 =
3
4
3
−2
3
=> 𝑝𝑜𝑖𝑛𝑡: 𝑝2 ( , 0,0)
4
_______________________________
5
−2
=> 𝑝𝑜𝑖𝑛𝑡: 𝑝1 ( , 0,0)
3
𝑝1 𝑝2 = 𝑝2 − 𝑝1 =<
⃗⃗⃗⃗⃗⃗⃗⃗⃗
17
, 0,0 >
12
_______________________________
Using ⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑝1 𝑝2 and 𝑛1 : 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑝1 𝑝2
𝑐𝑜𝑚𝑝⃗⃗⃗⃗⃗
𝑛1
=
|𝑝
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗1 |
1 𝑝2 .𝑛
‖𝑛
⃗⃗⃗⃗⃗1 ‖
=
17
2
√62 +32 +92
=
17
2√126
_______________________________
Or using ⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑝1 𝑝2 and 𝑛2 : 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑝1 𝑝2
𝑐𝑜𝑚𝑝⃗⃗⃗⃗⃗
𝑛2
=
|𝑝
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗2 |
1 𝑝2 .𝑛
‖𝑛
⃗⃗⃗⃗⃗2 ‖
=
17
3
√4 2 +22 +62
=
17
2√126
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6
3.4 Distance between two skewed lines:
To find the distance between two skewed lines:
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
1. Find vectors 𝑷
𝟏 𝑸𝟏 from the first line and 𝑷𝟐 𝑸𝟐 from the
second line.
2. Find vector ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑷𝟏 𝑷𝟐
3. 𝑫𝒊𝒏𝒔𝒕𝒂𝒏𝒄𝒆 =
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
|(𝑷
𝟏 𝑸𝟏 ×𝑷
𝟐 𝑸𝟐 ).𝑷
𝟏 𝑷𝟐 |
||𝑷𝟏 𝑸𝟏 ×𝑷𝟐 𝑸𝟐 ||
__________________________________________________________________
Example:
Find the shortest distance between the skewed lines:
𝒍𝟏 : 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑷𝟏 (𝟏, −𝟐, 𝟑), 𝑸𝟏 (𝟐, 𝟎, 𝟓)
𝒍𝟐 : 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑷𝟐 (𝟒, 𝟏, −𝟏), 𝑸𝟐 (−𝟐, 𝟑, 𝟒)?
Solution:
⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑃1 𝑄1 = 𝑄1 − 𝑃1 =< 1,2,2 >
⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑃2 𝑄2 = 𝑄2 − 𝑃2 =< −6,2,5 >
_______________________________
⃗⃗⃗⃗⃗⃗⃗⃗
𝑃1 𝑃2 = 𝑃2 − 𝑃1 =< 3,3, −4 >
_______________________________
𝑖
𝑗
⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝑃1 𝑄1 × 𝑃2 𝑄2 = | 1 2
−6 2
𝑘
2| = 6𝑖 − 17𝑗 + 14𝑘 =< 6, −17,14 >
5
_______________________________
7
𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 =
⃗⃗⃗⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗
|< 6, −17,14 >. < 3,3, −4 >|
|(𝑃
89
1 𝑄1 × 𝑃2 𝑄2 ). 𝑃1 𝑃2 |
=
=
||𝑃1 𝑄1 × 𝑃2 𝑄2 ||
√62 + 172 + 142
√521
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8