Distance from a Point to a line and plane 2012.notebook
June 03, 2014
9.5 Finding the Distance From a Point to a Line
y
n = (A, B)
Q(x1 , y )
Let P0(x0, y0) be any point on the line l. 1
d
Since RQ l, it is the direction of the normal vector n = (A, B) to the line l.
R
Then the distance d equals the length of the scalar projection of PQ on n.
P0 (x0 , y0 )
x
l : Ax + By + C =0
PQ • n
∴ d = n
(x1 ­ x0, y1 ­ y0) • (A, B)
= A2 + B2
A(x1 ­ x0) + B(y1 ­ y0)
= Since P0(x0, y0) lies on the line l, it satisfies the equation Ax + By + C = 0 or Axo + Byo + C = 0 or C = ­Axo ­ Byo.
A2 + B2
Ax 1 + By1 + (­Ax0 ­ By0)
= A2 + B2
The distance from the point Q(x, y) to the line Ax + By + C = 0 is: d = Ax1 + By1 + C
A2 + B2
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Ex 1. Find the distance from the point Q(2, ­3) to the line 4x + 5y ­ 6 = 0.
d = Ax1 + By1 + C
A2 + B2
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Finding the Distance From a Point to a Line in Space (R3)
Find the distance from Q to the given line l.
l is a line passing through P and having direction vector d. Q is a point, not on the line and FQ is the distance from Q to the line. FQ l . θ is the angle between PQ and d .
FQ
sinθ =
PQ
FQ = PQ sinθ
From ΔPFQ, FQ = PQ sinθ
= PQ sinθ ×
d
d
= PQ d sin θ
d
But PQ d = ×
PQ d sin θ
PQ d ×
therefore FQ = d
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Example 2:
Find the distance from the point Q (1, ­ 2, ­ 3) to the line r = (3, 1, 0) + t (1, 1, 2).
PQ d ×
therefore FQ = d
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Example:
Find the distance between the parallel lines r = (­ 2, 2, 1) + t (7, 3, ­ 4) and r = (2, ­ 1, ­ 2) + u (7, 3, ­ 4).
PQ d ×
therefore FQ = d
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Finding the Distance between Skew Lines
A1
A1
l1
l1
D
l2
A2
l2
A2
In order to find the distance between skew lines l1 and l2 we take the cross product of direction vectors of the lines to produce a vector perpendicular to both lines (a normal to the lines). Then we choose points A1 and A2 on each line and project the position vector onto the normal vector .
The length D of this projection is the distance between the lines.
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Example: Find the distance between the skew lines
l1:
x = 1 + t
y = t
z = t
l2:
x = 2s
y=1-s
z=s
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
9.6 The Shortest Distance from a Point to a Plane
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Example 1: Find the distance from the point (1, 2, 3) to the plane 2x + y ­ 2z ­ 4 = 0.
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Example 2: Find the distance between the planes π1 : 2x + 2y ­ z ­ 3 = 0 and π2: 4x + 4y ­ 2z + 9 = 0.
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Example 3: Find the foot of the perpendicular from
Q (2, 0, ­ 1) to the plane π: 3x ­ y ­ 2z ­ 1 = 0.
What is the shortest distance from the point Q to the plane???
Q(2, 0, ­1) R(
)
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Find the coordinates of the foot of the perpendicular from Q(3, 2, 4) to the line r = (­6, ­7, ­3) +t(5, 3, 4) then determine the shortest distance.
r = (­6, ­7, ­3) +t(5, 3, 4)
Q(3, 2, 4)
P(­6 + 5t, ­7 + 3t,­3 + 4t)
Parametric equations of the line:
x = ­6 + 5t
y = ­7 + 3t
z = ­3 + 4t
The coordinates of any point on the line are therefore (x, y, z) = (­6 + 5t, ­7 + 3t,­3 + 4t)
QP = (­6 + 5t ­ 3, ­7 + 3t ­ 2,­3 + 4t ­ 4)
QP = (­ 9 + 5t, ­ 9 + 3t, ­7 + 4t)
Because QP is perpendicular to the line, the dot product would be zero.
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
Homework pg 540 #1 ­ 3, 5 ­ 8
pg 550 #2, 3a, 4 ­ 7
Review pg 552 #3­10,12­17,19
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
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Distance from a Point to a line and plane 2012.notebook
June 03, 2014
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