Calculus & Vectors Class 15 handout
Relationships Between Point, Lines and Planes (2)
Distance from a Point to a Line in ℝ𝟐
In 2D-space, we know that a line’s direction is determined by its _________________. We can
use this fact to find the formula for the distance between a point and a line when they are
restricted to 2D-space:
In 2D-space, we have a line represented by its Cartesian equation 𝑙: 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 and an
arbitrary point 𝑃1 with coordinates 𝑃1 : (𝑥1 , 𝑦1 ), the distance 𝑑 from 𝑃1 to 𝑙 is determined by:
𝒅=
|𝐀𝒙𝟏 + 𝐁𝒚𝟏 + 𝐂|
√𝑨𝟐 +𝑩𝟐
Ex 1. Find the distance between the point (7, 5) to each of the following lines:
a) 4𝑦 − 10𝑥 = 7
b) 𝑟⃗ = (1, 4) + 𝑡(3, −2), 𝑡 ∈ ℝ
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Calculus & Vectors Class 15 handout
Distance from a Point to a Line in ℝ𝟑
We cannot develop the same formula to determine the distance between a point and a
line in 3D-space, as there is no Cartesian equation for a line when we allow one more
dimension (recall that a line in 3D has infinitely many non-parallel normal vectors).
However, we can draw a similar diagram to find another approach:
In 3D-space, we have a line represented by its vector equation 𝑙: 𝑟⃗ = ⃗⃗⃗⃗
𝑟0 + 𝑡𝑢⃗⃗, 𝑡 ∈ ℝ where ⃗⃗⃗⃗
𝑟0 =
⃗⃗⃗⃗⃗⃗⃗⃗
𝑂𝑃𝑜, the position vector of a given point 𝑃𝑜(𝑥𝑜, 𝑦𝑜 , 𝑧𝑜 ) on the line and 𝑢⃗⃗ = (𝑢𝑥, 𝑢𝑦 , 𝑢𝑧) is a
direction vector of the line. The distance 𝑑 from some other point 𝑃1 with coordinates
𝑃1 (𝑥1 , 𝑦1 , 𝑧1 ) to 𝑙 is determined by:
𝒅=
| ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
𝐏𝟎 𝐏𝟏 × 𝒖
⃗⃗⃗|
|𝒖
⃗⃗⃗|
Ex 2. Find the distance between the point (5, 4, −1) to the line
𝑙: 𝑟⃗ = (1, −3, 4) + 𝑡(4, 1, 2), 𝑡 ∈ ℝ.
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Calculus & Vectors Class 15 handout
There is another way to find the distance between a point and a line. Let’s use this
second method to solve the same problem in example 2:
Distance from a Point to a Plane
A plane 𝝅 is given by its Cartesian equation 𝝅: 𝐴𝑥 + 𝐵𝑦 + 𝐶𝑧 + 𝐷 = 0 and an
arbitrary point 𝑃1 (𝑥1 , 𝑦1 , 𝑧1 ) in 3D-space as shown below:
𝑑 =
⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗
|𝑃
⃗⃗ |
0 𝑃1 ∙ 𝑛
|𝑛⃗⃗|
(Note: this is just _______________________.)
From there, we can derive the formula for the distance 𝑑 from
the point 𝑃1 to the plane 𝝅:
𝒅=
|𝐀𝒙𝟏 + 𝐁𝒚𝟏 + 𝐂𝒛𝟏 +𝑫|
√𝑨𝟐 +𝑩𝟐 +𝑪𝟐
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Calculus & Vectors Class 15 handout
Ex 3. Find the distance from the point (5, 5, 10) to the plane 𝜋: 5𝑥 + 𝑦 − 6𝑧 = 60.
Ex 4. Find the distance between the parallel planes 𝜋1 : 3𝑥 + 2𝑦 − 4𝑧 = 30 and
𝜋2 : 6𝑥 + 4𝑦 − 8𝑧 + 7 = 0.
Intersection of Three Planes
Last lesson, we have seen that there are three possibilities for the intersection of two planes:
How many ways can three planes intersect each other? Assuming the three planes are not
parallel or coincident (or two of them coincident and 1 parallel). Then the planes could
possibly intersect:
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Calculus & Vectors Class 15 handout
We can see that out of all the possibilities, there is only one case (case 6) where all
three planes intersect at a point. If the three planes have a single point of intersection,
then when we solve the corresponding _______________________ representing those
three planes, we will arrive at a single solution. We call a system of three equations and
3 variables as such consistent. When we can’t obtain a single solution from a system
of three equations and 3 variables, we say that the system is inconsistent.
Ex 5. The following three Cartesian equation represent three planes that will intersect at
a point. Determine the point of intersection of these three planes.
𝜋1 : 3𝑥 − 𝑦 + 𝑧 + 7 = 0
𝜋2 : 2𝑥 − 3𝑦 + 𝑧 + 6 = 0
𝜋3 : 2𝑥 − 4𝑦 + 2𝑧 + 11 = 0
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Calculus & Vectors Class 15 handout
Think: Can you change one equation in example 5 so that the three planes intersect in
the way as shown in
a) case 6?
b) case 8?
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