MCV 4U3 – CALCULUS AND VECTORS Unit 8 – Relationships Between Points, Lines, and Planes

Lesson 45 – The Distance From a Point to a Line

Determining The Distance From A Point To A Line In R2: Example 1: Find the distance from the point Q(4, –8) to the line 6x + 3y – 10 = 0. Solution:

From the scalar equation of the line, the normal is (6, 3). The required distance is QN, where N is the point where the normal through Q meets the given line 6x + 3y – 10 = 0. QN is the magnitude of the projection of P0Q onto the normal to the line, where P0 is any point on the line.

Let P0 = (0, 310 )

Then P0Q = (4, 334

− )

4

2

-­‐2

-­‐4

-­‐6

-­‐8

-­‐10

-­‐5 5 10 15

6x + 3y – 10 = 0

Q(4, -8) N

QN = |Proj(P0Q onto n)|

= n

nQP •0

= ( )

( ) ( )22 36

3,6334,4

+

•⎟⎠

⎞⎜⎝

⎛ −

= 453424 −

= 5310−

= ( )53510−

= 352

The distance from the point Q(4, –8) to the line 6x + 3y – 10 = 0 is 352 units.

In general terms we can find a simple formula for the distance from a point Q(x1, y1) to a line with scalar equation Ax + By + C = 0. Letting P0(x0, y0) be a point on the line, the distance, d, is: d = |Proj(P0Q onto n)|

= n

nQP •0

= ( ) ( )22

0101 ,,

BA

BAyyxx

+

•−−

= 22

0101

BA

ByByAxAx

+

−+−

Since P(x0, y0) is on this line, it satisfies this line, so Ax0 + By0 + C = 0 or C = –Ax0 – By0

Then the distance is d = 22

11

BA

CByAx

+

++

Example 2: Find the distance from the point Q(4, –8) to the line 6x + 3y – 10 = 0

using the formula above. Solution: A = 6, B = 3, C = –10, 𝑥! = 4, and 𝑦! = −8

𝑑 =6 4 + 3 −8 − 10

6 ! + 3 !

𝑑 =−1045

𝑑 =103 5

𝑑 =103 5

55

𝑑 =10 515

𝑑 =2 53

The distance from the point Q(4, –8) to the line 6x + 3y – 10 = 0 is 352 units.

Determining The Distance From A Point To A Line In R3: It is not possible to use the formula we just developed for finding the distance between a point and a line in 3-space because you cannot write the equation of a line in 3-space in Cartesian form. We need to find another way. There are two methods: 1) The Cross Product:

Consider the following. We want to find the shortest distance from point P to the line 𝑟 = 𝑟! + 𝑡𝑑 where 𝑑 is the direction vector of the line. Let D represent the shortest distance from point P to the line 𝑟 = 𝑟! + 𝑡𝑑. Let θ represent the angle between QP and QR. In ΔPQR, sin𝜃 = !

!"

Therefore, 𝐷 = 𝑄𝑃 sin𝜃 𝑑×𝑄𝑃 = 𝑑 𝑄𝑃 sin𝜃 but 𝐷 = 𝑄𝑃 sin𝜃 𝑑×𝑄𝑃 = 𝑑 𝐷

If we isolate D we get,

𝐷 =𝑑×𝑄𝑃𝑑

Therefore, In R3, the distance from a point P to the line 𝑟 = 𝑟! + 𝑡𝑑 is:

𝐷 =𝑑×𝑄𝑃𝑑

Where Q is a point on the given line and 𝑑 is the direction vector of the given line.

P

Q R 𝑑

𝑟 = 𝑟!!!!⃗ + 𝑡𝑑

D

θ

Example 3: Determine the distance from P(0, –1, 0) to 𝑟 = 2, 1, 0 + 𝑡 −4, 5, 20 Solution: From the vector equation of the line, 𝑄 = (2, 1, 0) and 𝑑 = −4, 5, 20 . 𝑄𝑃 = 0,−1, 0 − 2, 1, 0 = (−2,−  2, 0)

𝐷 =−4, 5, 20 × −2,−2, 0

−4 ! + 5 ! + 20 !

𝐷 =40,−40, 18

441

𝐷 =40 ! + −40 ! + 18 !

21

𝐷 =352421

𝐷 =2 88121

Therefore the distance from P to the line is ! !!"

!" units.

2) The Dot Product:

We can redo the above example using the dot product method. We start by writing 𝑟 = 2, 1, 0 + 𝑡 −4, 5, 20 in parametric form. 𝑥 = 2− 4𝑡 𝑦 = 1+ 5𝑡 𝑧 = 20𝑡 Then we create a vector from a general point on the line 𝑄 2− 4𝑡, 1+ 5𝑡, 20𝑡 to 𝑃 0,−1, 0 . 𝑄𝑃 = 0− 2− 4𝑡 ,−1− 1+ 5𝑡 , 0− 20𝑡 𝑄𝑃 = −2+ 4𝑡,−2− 5𝑡,−20𝑡 We want to find the minimum distance from P to the line. This occurs when 𝑄𝑃 is perpendicular to the given line, or when 𝑑 ∙ 𝑄𝑃 = 0. −4, 5, 20 ∙ −2+ 4𝑡,−2− 5𝑡,−20𝑡 = 0 8− 16𝑡 − 10− 25𝑡 − 400𝑡 = 0 −2− 441𝑡 = 0 −441𝑡 = 2

𝑡 = −2441

So the minimal distance occurs when 𝑡 = − !

!!". At this value of t, 𝑄𝑃 is:

𝑄𝑃 = −2+ 4 −2441 ,−2− 5 −

2441 ,−20 −

2441

𝑄𝑃 = −890441 ,−

872441 ,

40441

The shortest distance is 𝑄𝑃

𝑄𝑃 = −890441

!

+ −872441

!

+40441

!

𝑄𝑃 =1554084194481

𝑄𝑃 =3524441

𝑄𝑃 =2 88121

Therefore the distance from P to the line is ! !!"

!" units.

This gives the same answer that we found using method 1. It has the advantage that it also allows us to find the coordinates of the point Q on the line that produces the minimal distance. Homework: Pages 540 – 541 # 1b # 2b # 5b # 6a # 8