Topic: Vectors equation of line in 3D, Distance from a point to a line
in 3D
Food for thought
• What is 𝑥-intercept?
• What is 𝑦-intercept?
Line in 2D (plane) and in 3D (Space)
A Line in plane (2D):
A line in the 𝒙𝒚-plane is determined when 𝒚-coordinate of a point (that is yintercept) on the line and the direction of the line (its slope or angle of
inclination) are given. The equation of the line can then be written using the
point-slope form.
Recall: Point-slope form:
For any point (𝒙𝟎 , 𝒚𝟎 ) on the line having slope 𝒎 is given by the equation
𝒚 − 𝒚𝟎 = 𝒎(𝒙 − 𝒙𝟎 )
If we fix the point, to be any point on the 𝒚-axis (𝒚-intercept), then 𝒙𝟎 = 𝟎
and 𝒚𝟎 = 𝒄, ∀ 𝒄 ∈ ℝ and above equation becomes
𝒚 − 𝒄 = 𝒎𝒙
or
𝒚 = 𝒎𝒙 + 𝒄
A Line in a Space (3D):
A line 𝑳 in three-dimensional space is determined when we know a point
𝑷𝟎 (𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 ) on 𝑳 and the direction of line 𝑳. In three dimensions, the direction
⃗ be a vector parallel to
of a line is conveniently described by a vector, so we let 𝒗
the line 𝑳.
Vector Equation for a Line in space:
Let 𝑷(𝒙, 𝒚, 𝒛) be an arbitrary point on 𝑳 and
⃗⃗⃗⃗𝟎 and 𝒓
⃗ be the position vectors of 𝑷𝟎 and
let 𝒓
𝑢
⃗
𝑷 (that is, they have representations ⃗⃗⃗⃗⃗⃗⃗⃗
𝑶𝑷𝟎 and
⃗⃗⃗⃗⃗⃗
⃗ is the vector with representation
𝑶𝑷). If 𝒖
⃗⃗⃗⃗⃗⃗⃗⃗
𝑷𝟎 𝑷, as in Figure 1, then the Triangle Law
for vector addition gives
Figure 1
⃗ =𝒓
⃗⃗⃗⃗𝟎 + 𝒖
⃗
𝒓
⃗ and 𝒗
⃗ are parallel vectors, there is a scalar 𝒕 ∈ (−∞, +∞) such that
But, since 𝒖
⃗ = 𝒕𝒗
⃗ . Thus
𝒖
⃗ =𝒓
⃗⃗⃗⃗𝟎 + 𝒕𝒗
⃗ _____________(𝟏)
𝒓
which is a vector equation of a line 𝑳.
Each value of the parameter 𝒕 gives the
⃗ of a point on 𝑳. In other
position vector 𝒓
words, as 𝒕 varies, the line is traced out by
⃗ . As Figure 2
the tip of the vector 𝒓
indicates, positive values of 𝒕 correspond
to points on 𝑳 that lie on one side of 𝑷𝟎 .
Figure 2
⃗ that gives the direction of the line 𝑳 is written in component form as
If the vector 𝒗
⃗ =< 𝒂, 𝒃, 𝒄 >, then we have 𝒕𝒗
⃗ =< 𝒕𝒂, 𝒕𝒃, 𝒕𝒄 >.
𝒗
⃗ =< 𝒙, 𝒚, 𝒛 >and 𝒓
⃗⃗⃗⃗𝟎 =< 𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 >, so the vector equation
We can also write 𝒓
becomes
< 𝒙, 𝒚, 𝒛 > = < 𝒙𝟎 + 𝒕𝒂, 𝒚𝟎 + 𝒕𝒃, 𝒛𝟎 + 𝒕𝒄 >
Two vectors are equal if and only if corresponding components are equal.
Therefore, we have the three scalar equations:
𝒙 = 𝒙𝟎 + 𝒕𝒂,
𝒚 = 𝒚𝟎 + 𝒕𝒃,
𝒛 = 𝒛𝟎 + 𝒕𝒄 _________________________(𝟐)
where 𝒕 ∈ ℝ.
These equations are called parametric equations of the line 𝑳 through the point
𝑷𝟎 (𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 ) and parallel to the vector ⃗𝒗 =< 𝒂, 𝒃, 𝒄 >. Each value of the
parameter 𝒕 gives a point (𝒙, 𝒚, 𝒛) on 𝑳.
Parametric Equation for a Line in space
Parametric equations for a line through the point 𝑷𝟎 (𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 ) and parallel to
⃗ =< 𝒂, 𝒃, 𝒄 > are
the direction vector 𝒗
𝒙 = 𝒙𝟎 + 𝒕𝒂
𝒚 = 𝒚𝟎 + 𝒕𝒃
𝒛 = 𝒛𝟎 + 𝒕𝒄
Note:
⃗ and 𝒗
⃗ are parallel if 𝒖
⃗ =𝒕𝒗
⃗ , where 𝒕 is scalar. We understand it
Two vectors 𝒖
with the help of an example given below.
⃗ = < 𝟏, 𝟐, 𝟑 >
𝒖
⃗ = < 𝟐, 𝟒, 𝟔 >
𝒗
Then
⃗ = 𝟐 < 𝟏, 𝟐, 𝟑 >
𝒗
⃗ = 𝟐𝒖
⃗
𝒗
⃗ &𝒗
⃗ are parallel.
This implies that 𝒖
Example 1:
Find parametric equation & vector equation for the line through the point
(−𝟐, 𝟎, 𝟒) and parallel to the vector
⃗
⃗ = 𝟐𝒊 + 𝟒𝒋 − 𝟐𝒌
𝒗
Solution:
Since the given point is
𝑃 = (𝑥0 , 𝑦0 , 𝑧0 ) = (−2,0,4)
and the vector is
⃗ = 𝑣1 𝑖 + 𝑣2 𝑗 + 𝑣3 𝑘
⃗ = 2𝑖 + 4𝑗 − 2𝑘
⃗
𝑣 = a𝑖 + b𝑗 + c𝑘
Here we have
𝑥0 = −2,
𝑎 = 2,
𝑦0 = 0,
𝑏 = 4,
𝑧0 = 4
𝑐 = −2
Parametric equation of line in space:
𝑥 = 𝑥𝑜 + 𝑡𝑎 = −2 + 2𝑡
𝑦 = 𝑦𝑜 + 𝑡𝑏 = 0 + 4𝑡 = 4𝑡
𝑧 = 𝑧𝑜 + 𝑡𝑐 = 4 − 2𝑡
That is,
𝑥 = −2 + 2𝑡
𝑦 = 4𝑡
𝑧 = 4 − 2𝑡
where −∞ < 𝒕 < ∞.
Vector Equation of Line in space:
𝑟 = 𝑟0 + 𝑡 𝑣
⃗ = (−2𝑖 + 0𝑗 + 4𝑘
⃗ ) + 𝑡 (2𝑖 + 4𝑗 − 2𝑘
⃗ ) ; −∞ < 𝑡 < ∞
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
⃗ = (−2⃗𝑖 + 0⃗𝑗 + 4⃗⃗⃗𝑘) + (2𝑡 𝑖 + 4𝑡 𝑗 − 2𝑡 𝑘
⃗)
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
⃗ = (−2 + 2𝑡) 𝑖 + (0 + 4𝑡)𝑗 + (4 − 2𝑡) 𝑘
⃗
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
⃗ = (−2 + 2𝑡) 𝑖 + (4𝑡)𝑗 + (4 − 2𝑡) 𝑘
⃗
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘
Example 2:
Find the parametric equation of the line passing through origin and parallel to the
vector
𝑣 = 2𝑗̂ + 𝑘̂
Solution:
Given that
Origin= 𝑃 = (0,0,0) = 𝑃(𝑥0 , 𝑦0 , 𝑧0 )
𝑣 = 0𝑖̂ + 2𝑗̂ + 𝑘̂ = 𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂
Then the parametric equations of line are
𝑥 = 𝑥0 + 𝑎𝑡 = 0 + 0𝑡 = 0
𝑦 = 𝑦0 + 𝑏𝑡 = 0 + 2𝑡 = 2𝑡
𝑧 = 𝑧0 + 𝑐𝑡 = 0 + 𝑡 = 𝑡
Then the required parametric equations are
𝒙=𝟎
𝒚 = 𝟐𝒕
𝒛=𝒕
Example 3:
Find the parametric equation and vector equation of the line through
𝑷(−𝟑, 𝟐, −𝟑) and 𝑸(𝟏, −𝟏, 𝟒).
Solution:
𝑃 = (𝑥0 , 𝑦0 , 𝑧0 ) = (−3,2, −3)
𝑄 = (𝑥1 , 𝑦1 , 𝑧1 ) = (1, −1,4)
The vector parallel to line is
⃗⃗⃗⃗⃗ =< 𝑥1 − 𝑥0 ,
𝑣 = 𝑃𝑄
𝑦1 − 𝑦0 ,
𝑧1 − 𝑧0 >
⃗⃗⃗⃗⃗ =< 1 − (−3),
𝑣 = 𝑃𝑄
−1 − 2,
4 − (−3) >
⃗⃗⃗⃗⃗ =< 4, −3, 7 > =< 𝑎, 𝑏, 𝑐 >
𝑣 = 𝑃𝑄
We can write the vector in standard form as
⃗ = 𝟒𝒊 − 𝟑𝒋 + 𝟕𝒌
⃗
⃗ = 𝒂𝒊 + 𝒃𝒋 + 𝒄𝒌
𝒗
Parametric equation of a line in space:
𝑥 = 𝑥𝑜 + 𝑡 𝑣1 = −3 + 4𝑡
𝑦 = 𝑦𝑜 + 𝑡 𝑣2 = 2 − 3𝑡
𝑧 = 𝑧𝑜 + 𝑡 𝑣3 = −3 + 7𝑡
where −∞ < 𝑡 < ∞.
Vector equation for a line in space:
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = (−3𝑖 + 2𝑗 − 3𝑘) + 𝑡 (4𝑖 − 3𝑗 + 7𝑘) , −∞ < 𝑡 < ∞
𝑥𝑖 + 𝑦𝑗 + 𝑧𝑘 = (−3𝑖 + 2𝑗 − 3𝑘) + (4𝑡 𝑖 − 3𝑡 𝑗 + 7𝑡 𝑘)
𝒙𝒊 + 𝒚𝒋 + 𝒛𝒌 = (−𝟑 + 𝟒𝒕) 𝒊 + (𝟐 − 𝟑𝒕)𝒋 + (−𝟑 + 𝟕𝒕)𝒌
Do yourself Exercise 12.5 (Textbook: Thomas Calculus 11th Edition):
Q # 1 to 20.
Example 4:
Find the parametric equation of the line passing through the point (3, −2,1) and
parallel to the line having parametric equations
𝑥 = 1 + 2𝑡
𝑦 =2−𝑡
𝑧 = 3𝑡
Solution:
Since we have given the parametric equations as
𝑥 = 1 + 2𝑡 = 𝑥0 + 𝑎𝑡
𝑦 = 2 − 𝑡 = 𝑦0 + 𝑏𝑡
𝑧 = 0 + 3𝑡 = 𝑧0 + 𝑐𝑡
This implies that
𝑎 = 2,
𝑏 = −1,
𝑐=3
which are the components of a vector 𝒗, given as
𝒗 =< 𝒂, 𝒃, 𝒄 >=< 𝟐, −𝟏, 𝟑 >
The given point is
𝑃 = (3, −2,1) = 𝑃(𝑥0 , 𝑦0 , 𝑧0 )
𝒗 =< 𝒂, 𝒃, 𝒄 ≥< 𝟐, −𝟏, 𝟑 >
That is,
𝑥0 = 3,
𝑦0 = −2,
𝑧0 = 1
𝑎 = 2,
𝑏 = −1,
𝑐=3
Then the parametric equation of the line passing through the point 𝑃(3, −2,1)
and parallel to the line with the given parametric equations are given below:
𝑥 = 𝑥0 + 𝑎𝑡
𝑦 = 𝑦0 + 𝑏𝑡
𝑧 = 𝑧0 + 𝑐𝑡
Putting the values, we have
𝒙 = 𝟑 + 𝟐𝒕
𝒚 = −𝟐 − 𝒕
𝒛 = 𝟏 + 𝟑𝒕
which are the required parametric equations of the line.
Example 5:
Find the parametric equations of the line passing through the point (2, 3, 0), and
perpendicular to the vectors 𝑢
⃗ = 𝑖̂ + 2𝑗̂ + 3𝑘̂ and 𝑣 = 3𝑖̂ + 4𝑗̂ + 5𝑘̂.
Solution:
Since the line is perpendicular to the vectors 𝑢
⃗ and 𝑣, therefore using the cross
product, we have
𝑖̂ 𝑗̂ 𝑘̂
𝑢
⃗ × 𝑣 = |1 2 3 |
3 4 5
𝑢
⃗ × 𝑣 = 𝑖̂(10 − 12) − 𝑗̂(5 − 9) + 𝑘̂(4 − 6)
𝑢
⃗ × 𝑣 = −2𝑖̂ + 4𝑗̂ − 2𝑘̂ = 𝑎𝑖̂ + 𝑏𝑗̂ − 𝑐𝑘̂
The given point is
𝑃 = (2,3,0) = 𝑃(𝑥0 , 𝑦0 , 𝑧0 )
Then the parametric equations of line are
𝑥 = 𝑥0 + 𝑎𝑡 = 2 − 2𝑡
𝑦 = 𝑦0 + 𝑏𝑡 = 3 + 4𝑡
𝑧 = 𝑧0 + 𝑐𝑡 = 0 − 2𝑡
Then the required parametric equations are
𝑥 = 2 − 2𝑡
𝑦 = 3 + 4𝑡
𝑧 = −2𝑡
Practice questions:
(Textbook: Thomas Calculus 11th Edition) Ex. 12.5: 1-7, 10.
Parameterization of a Line segment
Example:
Parameterize the line segment joining the points 𝑷 (−𝟑, 𝟐, −𝟑) and 𝑸 (𝟏, −𝟏, 𝟒).
Solution:
First of all, we will find the parametric equation of the line through the points
𝑷 (−𝟑, 𝟐, −𝟑) and 𝑸 (𝟏, −𝟏, 𝟒) and then restrict the domain of parameter 𝒕 to obtain
the parametric equation of the line segment from 𝑷 to 𝑸.
Step 1: (Equation of line)
⃗⃗⃗⃗⃗ = (1 + 3)𝑖̂ + (−1 − 2)𝑗̂ + (4 + 3)𝑘̂
𝑣 = 𝑃𝑄
̂ = 𝒂𝒊̂ + 𝒃𝒋̂ + 𝒄𝒌
̂
⃗ = 𝟒𝒊̂ − 𝟑𝒋̂ + 𝟕𝒌
𝒗
Let us consider the point 𝑷 (−𝟑, 𝟐, −𝟑) = 𝑷(𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 ). (NOTE: We can also
consider point Q here). Then the parametric equations of line are
𝒙 = 𝒙𝟎 + 𝒂𝒕 = −𝟑 + 𝟒𝒕
𝒚 = 𝒚𝟎 + 𝒃𝒕 = 𝟐 − 𝟑𝒕
𝒛 = 𝒛𝟎 + 𝒄𝒕 = −𝟑 + 𝟕𝒕
Step 2: (Line segment)
In order to find the value of 𝑡 for which an arbitrary point (𝒙, 𝒚, 𝒛) of the line is at
𝑷 (−𝟑, 𝟐, −𝟑), we solve the equation
−3 = −3 + 4𝑡
}⇒𝑡=0
2 = 2 − 3𝑡
−3 = −3 + 7𝑡
Similarly, when (𝒙, 𝒚, 𝒛) is at 𝑸 (𝟏, −𝟏, 𝟒), we solve
1 = −3 + 4𝑡
−1 = 2 − 3𝑡} ⇒ 𝑡 = 1
4 = −3 + 7𝑡
So, the parametric equation of the line segment is
𝑥 = −3 + 4𝑡
𝑦 = 2 − 3𝑡
𝑧 = −3 + 7𝑡;
0≤𝑡≤1
Question 19: Find the parametric equations of the line segment joining the points
𝑷(−𝟐, 𝟎, 𝟐) and 𝑸(𝟎, 𝟐, 𝟎).
Practice Questions
(Textbook: Thomas Calculus 11th Edition) Ex. 12.5: 13-20.
The Distance from a Point to a Line in Space
⃗ be a vector parallel
Let 𝑳 be a line and let 𝒗
⃗ ||𝑳.
to the line 𝑳. This means that 𝒗
Let 𝑷𝟎 be a point in space from which we
need to measure the distance to the line 𝑳,
i.e., we need to measure the distance between
𝑷𝟎 and 𝑳.
We consider an arbitrary point 𝑷𝟏 on the line 𝑳 assuming it might give us a distance
from the point 𝑷𝟎 .
⃗ 𝟎 and 𝒓
⃗ 𝟏 respectively from
Considering 𝑷𝟎 and 𝑷𝟏 , we draw their position vectors 𝒓
the origin.
⃗ at an angle 𝜽 from 𝑷𝟏 to 𝑷𝟎 such that by head to tail rule we
We draw a vector 𝒖
have,
⃗𝟏+𝒖
⃗ =𝒓
⃗𝟎
𝒓
This implies that
⃗ = ⃗𝒓𝟎 − ⃗𝒓𝟏 .
𝒖
Now logically the shortest distance of a point from a line is the length of the
perpendicular drawn from the point to the line.
Let us consider a point 𝑷𝟐 on the line 𝑳, such that 𝑷𝟎 𝑷𝟐 is perpendicular to the line
̅̅̅̅̅̅̅
𝑳. i.e., 𝑷𝟎 𝑷𝟐 ⊥ 𝑳. We say that 𝑷
𝟎 𝑷𝟐 = 𝑠.
Distance is a scalar quantity. From the figure, we can write
𝒔𝒊𝒏 𝜽 =
𝑷𝒆𝒓𝒑𝒂𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓
𝒔
=
𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆
⃗⃗⃗⃗𝟎 − 𝒓
⃗⃗⃗⃗𝟏 |
|𝒓
𝒔𝒊𝒏 𝜽 =
𝒔
⃗|
|𝒖
|𝒖
⃗ | 𝒔𝒊𝒏 𝜽 = 𝒔
⃗ | 𝒔𝒊𝒏 𝜽
𝒔 = |𝒖
⃗ and 𝒗
⃗ . 𝑷𝟎 𝑷𝟐 ⊥ 𝑳 and 𝒗
⃗ ||𝑳 implies that 𝑷𝟎 𝑷𝟐 ⊥ 𝒗
⃗.
Here 𝜽 is the angle between 𝒖
⃗ and 𝒖
⃗ , we have
Now considering 𝒗
⃗ × 𝒗
⃗ = |𝒖
⃗ | |𝒗
⃗ | 𝐬𝐢𝐧 𝜽
𝒖
We can rearrange as
⃗ × 𝒗
⃗ = |𝒗
⃗ ||𝒖
⃗ | 𝒔𝒊𝒏 𝜽
𝒖
⃗ |𝒔𝒊𝒏 𝜽
We know 𝒔 = |𝒖
⃗ × 𝒗
⃗ = |𝒗
⃗ |𝒔
𝒖
or
⃗ × 𝒗
⃗ = 𝒔 |𝒗
⃗|
𝒖
Since distance is a scalar quantity thus, we need to take the magnitude of the cross
product to balance the equation.
|𝒖
⃗ ×𝒗
⃗ | = 𝒔 |𝒗
⃗|
|𝒖
⃗ ×𝒗
⃗|
=𝒔
|𝒗
⃗|
This implies that the distance between the line 𝑳 and a point can be computed by
the formula,
𝒔=
|𝒖
⃗ ×𝒗
⃗|
|𝒗
⃗|
⃗ =𝒓
⃗⃗⃗⃗𝟎 − 𝒓
⃗⃗⃗⃗𝟏
Where 𝒖
Example:
Find the distance from the point 𝑷𝟏 (𝟏, 𝟏, 𝟓) to the line:
𝒙=𝟏+𝒕
𝑳: {𝒚 = 𝟑 − 𝒕
𝒛 = 𝟐𝒕
Solution:
Since the given point is 𝑷𝟏 (𝒙𝟏 , 𝒚𝟏 , 𝒛𝟏 ) = 𝑷𝟏 (𝟏, 𝟏, 𝟓).
And the given parametric equations of line are
𝒙 = 𝒙𝟎 + 𝒂𝒕 = 𝟏 + 𝒕
𝒚 = 𝒚𝟎 + 𝒃𝒕 = 𝟑 − 𝒕
𝒛 = 𝒛𝟎 + 𝒄𝒕 = 𝟎 + 𝟐𝒕
⃗ =< 𝒂, 𝒃, 𝒄 > =< 𝟏, −𝟏, 𝟐 >.
This means that 𝑷𝟎 (𝒙𝟎 , 𝒚𝟎 , 𝒛𝟎 ) = 𝑷𝟎 (𝟏, 𝟑, 𝟎) and 𝒗
Thus, the vector parallel to the line 𝑳 is
̂
⃗ = 𝒊̂ − 𝒋̂ + 𝟐𝒌
𝒗
The Line passes through the point 𝑷𝟎 (𝟏, 𝟑, 𝟎)
𝑢
⃗ = ⃗⃗⃗⃗⃗⃗⃗⃗
𝑃0 𝑃1 = ⃗⃗⃗⃗⃗⃗⃗
𝑂𝑃1 − ⃗⃗⃗⃗⃗⃗⃗
𝑂𝑃0
𝑢
⃗ = ⃗⃗⃗⃗⃗⃗⃗⃗
𝑃0 𝑃1 = (𝑖̂ + 𝑗̂ + 5𝑘̂) − (𝑖̂ + 3𝑗̂ + 0 ̂𝑘)
𝑢
⃗ = (1 − 1)𝑖̂ + (1 − 3)𝑗̂ + (5 − 0)𝑘̂
𝑢
⃗ = 0𝑖̂ − 2𝑗̂ + 5𝑘̂.
Now,
𝑖̂
𝑗̂ 𝑘̂
𝑢
⃗ × 𝑣 = |0 −2 5|
1 −1 2
𝑢
⃗ × 𝑣 = 𝑖̂ |
0 −2
−2 5
0 5
| − 𝑗̂ |
| + 𝑘̂ |
|
1 −1
−1 2
1 2
𝑢
⃗ × 𝑣 = 𝑖̂(−4 + 5) − 𝑗̂(0 − 5) + 𝑘̂(0 + 2)
𝑢
⃗ × 𝑣 = 𝑖̂ + 5𝑗̂ + 2𝑘̂
⃗ ×𝒗
⃗ , we have
Taking magnitude of 𝒖
|𝑢
⃗ × 𝑣| = √(1)2 + (5)2 + (2)2
|𝑢
⃗ × 𝑣| = √30
⃗ , we have
Similarly, taking magnitude of vector 𝒗
|𝑣| = √(1)2 + (−1)2 + (2)2 = √6
Now, apply the formula, we have
𝑑=
|𝑢
⃗ × 𝑣|
|𝑣 |
Putting values, we have
𝑑=
√30
√6
=√
30
= √5
6
𝑑 = √5 = 2.24
Hence the distance of a point 𝑷𝟏 to the line 𝑳 is 2.24 units.
Practice Questions:
Textbook: Thomas Calculus 11th Edition Ex. 12.5: 33-38