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Слайды и текст к этой презентации:
№1 слайд
Содержание слайда:
№2 слайд
Содержание слайда: VECTORS AND THE GEOMETRY OF SPACE
A line in the xy-plane is determined when
a point 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.
№3 слайд
Содержание слайда:
№4 слайд
Содержание слайда: EQUATIONS OF LINES
A line L in three-dimensional (3-D) space
is determined when we know:
A point P0(x0, y0, z0) on L
The direction of L
№5 слайд
Содержание слайда: EQUATIONS OF LINES
In three dimensions, the direction
of a line is conveniently described by
a vector.
№6 слайд
Содержание слайда: EQUATIONS OF LINES
So, we let v be a vector parallel to L.
Let P(x, y, z) be an arbitrary point on L.
Let r0 and r be the position vectors of P0 and P.
That is, they have representations and .
№7 слайд
Содержание слайда: EQUATIONS OF LINES
If a is the vector with representation ,
then the Triangle Law for vector addition
gives:
r = r0 + a
№8 слайд
Содержание слайда: EQUATIONS OF LINES
However, since a and v are parallel vectors, there is a scalar t such that
a = tv
№9 слайд
Содержание слайда: VECTOR EQUATION OF A LINE
Thus,
r = r0 + t v
This is a vector equation of L.
№10 слайд
Содержание слайда: VECTOR EQUATION
Each value of the parameter t gives
the position vector r of a point on L.
That is, as t varies,
the line is traced
out by the tip of
the vector r.
№11 слайд
Содержание слайда: VECTOR EQUATION
Positive values of t correspond to points on L that lie on one side of P0.
Negative values correspond to points that
lie on the other side.
№12 слайд
Содержание слайда: VECTOR EQUATION
If the vector v that gives the direction of
the line L is written in component form as
v = <a, b, c>, then we have:
tv = <ta, tb, tc>
№13 слайд
Содержание слайда: VECTOR EQUATION
We can also write:
r = <x, y, z> and r0 = <x0, y0, z0>
So, vector Equation 1 becomes:
<x, y, z> = <x0 + ta, y0 + tb, z0 + tc>
№14 слайд
Содержание слайда: VECTOR EQUATION
Two vectors are equal if and only if corresponding components are equal.
Hence, we have the following three
scalar equations.
№15 слайд
Содержание слайда: SCALAR EQUATIONS OF A LINE
x = x0 + at
y = y0 + bt
z = z0 + ct
Where, t
№16 слайд
Содержание слайда: PARAMETRIC EQUATIONS
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.
№17 слайд
Содержание слайда: EQUATIONS OF LINES
Find a vector equation and parametric equations for the line that passes through
the point (5, 1, 3) and is parallel to the vector i + 4 j – 2 k.
Find two other points on the line.
№18 слайд
Содержание слайда: EQUATIONS OF LINES
Here, r0 = <5, 1, 3> = 5 i + j + 3 k
and v = i + 4 j – 2 k
So, vector Equation 1 becomes:
r = (5 i + j + 3 k) + t(i + 4 j – 2 k)
or
r = (5 + t) i + (1 + 4t) j + (3 – 2t) k
№19 слайд
Содержание слайда: EQUATIONS OF LINES
Parametric equations are:
x = 5 + t y = 1 + 4t z = 3 – 2t
№20 слайд
Содержание слайда: EQUATIONS OF LINES
Choosing the parameter value t = 1
gives x = 6, y = 5, and z = 1.
So, (6, 5, 1) is a point on the line.
Similarly, t = –1 gives the point (4, –3, 5).
№21 слайд
Содержание слайда: EQUATIONS OF LINES
The vector equation and parametric equations of a line are not unique.
If we change the point or the parameter
or choose a different parallel vector, then
the equations change.
№22 слайд
Содержание слайда: EQUATIONS OF LINES
For instance, if, instead of (5, 1, 3),
we choose the point (6, 5, 1) in Example 1,
the parametric equations of the line become:
x = 6 + t y = 5 + 4t z = 1 – 2t
№23 слайд
Содержание слайда: EQUATIONS OF LINES
Alternatively, if we stay with the point (5, 1, 3) but choose the parallel vector 2 i + 8 j – 4 k, we arrive at:
x = 5 + 2t y = 1 + 8t z = 3 – 4t
№24 слайд
Содержание слайда: DIRECTION NUMBERS
In general, if a vector v = <a, b, c> is used
to describe the direction of a line L, then
the numbers a, b, and c are called direction numbers of L.
№25 слайд
Содержание слайда: DIRECTION NUMBERS
Any vector parallel to v could also be used.
Thus, we see that any three numbers proportional to a, b, and c could also be used as a set of direction numbers for L.
№26 слайд
Содержание слайда: EQUATIONS OF LINES
Another way of describing a line L
is to eliminate the parameter t from Equations 2.
If none of a, b, or c is 0, we can solve each
of these equations for t, equate the results,
and obtain the following equations.
№27 слайд
Содержание слайда: SYMMETRIC EQUATIONS
These equations are called symmetric equations of L.
№28 слайд
Содержание слайда: SYMMETRIC EQUATIONS
Notice that the numbers a, b, and c that appear in the denominators of Equations 3
are direction numbers of L.
That is, they are components of a vector
parallel to L.
№29 слайд
Содержание слайда: SYMMETRIC EQUATIONS
If one of a, b, or c is 0, we can still eliminate t.
For instance, if a = 0, we could write
the equations of L as:
This means that L lies
in the vertical plane x = x0.
№30 слайд
Содержание слайда: EQUATIONS OF LINES
Find parametric equations and symmetric equations of the line that passes through the points A(2, 4, –3) and B(3, –1, 1).
At what point does this line intersect
the xy-plane?
№31 слайд
Содержание слайда: EQUATIONS OF LINES
We are not explicitly given a vector parallel
to the line.
However, observe that the vector v with representation is parallel to the line
and
v = <3 – 2, –1 – 4, 1 – (–3)> = <1, –5, 4>
№32 слайд
Содержание слайда: EQUATIONS OF LINES
Thus, direction numbers are:
a = 1, b = –5, c = 4
№33 слайд
Содержание слайда: EQUATIONS OF LINES
Taking the point (2, 4, –3) as P0,
we see that:
Parametric Equations 2 are:
x = 2 + t y = 4 – 5t z = –3 + 4t
Symmetric Equations 3 are:
№34 слайд
Содержание слайда: EQUATIONS OF LINES
The line intersects the xy-plane when z = 0.
So, we put z = 0 in the symmetric equations and obtain:
This gives x = and y = .
№35 слайд
Содержание слайда: EQUATIONS OF LINES
The line intersects the xy-plane
at the point
№36 слайд
Содержание слайда: EQUATIONS OF LINES
In general, the procedure of Example 2 shows that direction numbers of the line L through
the points P0(x0, y0, z0) and P1(x1, y1, z1)
are: x1 – x0 y1 – y0 z1 – z0
So, symmetric equations of L are:
№37 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
Often, we need a description, not of
an entire line, but of just a line segment.
How, for instance, could we describe
the line segment AB in Example 2?
№38 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
If we put t = 0 in the parametric equations
in Example 2 a, we get the point (2, 4, –3).
If we put t = 1, we get (3, –1, 1).
№39 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
So, the line segment AB is described by either:
The parametric equations
x = 2 + t y = 4 – 5t z = –3 + 4t
where 0 ≤ t ≤ 1
The corresponding vector equation
r(t) = <2 + t, 4 – 5t, –3 + 4t>
where 0 ≤ t ≤ 1
№40 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
In general, we know from Equation 1 that
the vector equation of a line through the (tip
of the) vector r0 in the direction of a vector v
is:
r = r0 + t v
№41 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
If the line also passes through (the tip of) r1, then we can take v = r1 – r0.
So, its vector equation is:
r = r0 + t(r1 – r0) = (1 – t)r0 + t r1
The line segment from r0 to r1 is given by
the parameter interval 0 ≤ t ≤ 1.
№42 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
The line segment from r0 to r1 is given by
the vector equation
r(t) = (1 – t)r0 + t r1
where 0 ≤ t ≤ 1
№43 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
Show that the lines L1 and L2 with parametric equations
x = 1 + t y = –2 + 3t z = 4 – t
x = 2s y = 3 + s z = –3 + 4s
are skew lines.
That is, they do not intersect and are not parallel, and therefore do not lie in the same plane.
№44 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
The lines are not parallel because
the corresponding vectors <1, 3, –1>
and <2, 1, 4> are not parallel.
Their components are not proportional.
№45 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
If L1 and L2 had a point of intersection,
there would be values of t and s such that
1 + t = 2s
–2 + 3t = 3 + s
4 – t = –3 + 4s
№46 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
However, if we solve the first two equations, we get:
t = and s =
These values don’t satisfy
the third equation.
№47 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
Thus, there are no values of t and s
that satisfy the three equations.
So, L1 and L2 do not intersect.
№48 слайд
Содержание слайда: EQUATIONS OF LINE SEGMENTS
Hence, L1 and L2 are skew lines.
№49 слайд
Содержание слайда: PLANES
Although a line in space is determined by
a point and a direction, a plane in space is more difficult to describe.
A single vector parallel to a plane is not enough
to convey the ‘direction’ of the plane.
№50 слайд
Содержание слайда: PLANES
However, a vector perpendicular
to the plane does completely specify
its direction.
№51 слайд
Содержание слайда: PLANES
Thus, a plane in space is determined
by:
A point P0(x0, y0, z0)
in the plane
A vector n that is
orthogonal to the plane
№52 слайд
Содержание слайда: NORMAL VECTOR
This orthogonal vector n is called
a normal vector.
№53 слайд
Содержание слайда: PLANES
Let P(x, y, z) be an arbitrary point in the plane.
Let r0 and r1 be the position vectors of P0
and P.
Then, the vector r – r0
is represented by
№54 слайд
Содержание слайда: PLANES
The normal vector n is orthogonal to every vector in the given plane.
In particular, n is orthogonal to r – r0.
№55 слайд
Содержание слайда: EQUATIONS OF PLANES
Thus, we have:
n . (r – r0) = 0
№56 слайд
Содержание слайда: EQUATIONS OF PLANES
That can also be written as:
n . r = n . r0
№57 слайд
Содержание слайда: VECTOR EQUATION
Either Equation 5 or Equation 6
is called a vector equation of the plane.
№58 слайд
Содержание слайда: EQUATIONS OF PLANES
To obtain a scalar equation for the plane,
we write:
n = <a, b, c>
r = <x, y, z>
r0 = <x0, y0, z0>
№59 слайд
Содержание слайда: EQUATIONS OF PLANES
Then, the vector Equation 5
becomes:
<a, b, c> . <x – x0, y – y0, z – z0> = 0
№60 слайд
Содержание слайда: SCALAR EQUATION
That can also be written as:
a(x – x0) + b(y – y0) + c(z – z0) = 0
This equation is the scalar equation
of the plane through P0(x0, y0, z0) with
normal vector n = <a, b, c>.
№61 слайд
Содержание слайда: EQUATIONS OF PLANES
Find an equation of the plane through
the point (2, 4, –1) with normal vector
n = <2, 3, 4>.
Find the intercepts and sketch the plane.
№62 слайд
Содержание слайда: EQUATIONS OF PLANES
In Equation 7, putting
a = 2, b = 3, c = 4, x0 = 2, y0 = 4, z0 = –1,
we see that an equation of the plane is:
2(x – 2) + 3(y – 4) + 4(z + 1) = 0
or
2x + 3y + 4z = 12
№63 слайд
Содержание слайда: EQUATIONS OF PLANES
To find the x-intercept, we set y = z = 0
in the equation, and obtain x = 6.
Similarly, the y-intercept is 4 and
the z-intercept is 3.
№64 слайд
Содержание слайда: EQUATIONS OF PLANES
This enables us to sketch the portion
of the plane that lies in the first octant.
№65 слайд
Содержание слайда: EQUATIONS OF PLANES
By collecting terms in Equation 7
as we did in Example 4, we can rewrite
the equation of a plane as follows.
№66 слайд
Содержание слайда: LINEAR EQUATION
ax + by + cz + d = 0
where d = –(ax0 + by0 + cz0)
This is called a linear equation
in x, y, and z.
№67 слайд
Содержание слайда: LINEAR EQUATION
Conversely, it can be shown that, if
a, b, and c are not all 0, then the linear Equation 8 represents a plane with normal vector <a, b, c>.
See Exercise 77.
№68 слайд
Содержание слайда: EQUATIONS OF PLANES
Find an equation of the plane that passes through the points
P(1, 3, 2), Q(3, –1, 6), R(5, 2, 0)
№69 слайд
Содержание слайда: EQUATIONS OF PLANES
The vectors a and b corresponding to and are:
a = <2, –4, 4> b = <4, –1, –2>
№70 слайд
Содержание слайда: EQUATIONS OF PLANES
Since both a and b lie in the plane,
their cross product a x b is orthogonal
to the plane and can be taken as the normal vector.
№71 слайд
Содержание слайда: EQUATIONS OF PLANES
Thus,
№72 слайд
Содержание слайда: EQUATIONS OF PLANES
With the point P(1, 2, 3) and the normal
vector n, an equation of the plane is:
12(x – 1) + 20(y – 3) + 14(z – 2) = 0
or
6x + 10y + 7z = 50
№73 слайд
Содержание слайда: EQUATIONS OF PLANES
Find the point at which the line with
parametric equations
x = 2 + 3t y = –4t z = 5 + t
intersects the plane
4x + 5y – 2z = 18
№74 слайд
Содержание слайда: EQUATIONS OF PLANES
We substitute the expressions for x, y, and z from the parametric equations into the equation of the plane:
4(2 + 3t) + 5(–4t) – 2(5 + t) = 18
№75 слайд
Содержание слайда: EQUATIONS OF PLANES
That simplifies to –10t = 20.
Hence, t = –2.
Therefore, the point of intersection occurs
when the parameter value is t = –2.
№76 слайд
Содержание слайда: EQUATIONS OF PLANES
Then,
x = 2 + 3(–2) = –4
y = –4(–2) = 8
z = 5 – 2 = 3
So, the point of intersection is (–4, 8, 3).
№77 слайд
Содержание слайда: PARALLEL PLANES
Two planes are parallel
if their normal vectors are
parallel.
№78 слайд
Содержание слайда: PARALLEL PLANES
For instance, the planes
x + 2y – 3z = 4 and 2x + 4y – 6z = 3
are parallel because:
Their normal vectors are
n1 = <1, 2, –3> and n2 = <2, 4, –6>
and n2 = 2n1.
№79 слайд
Содержание слайда: NONPARALLEL PLANES
If two planes are not parallel, then
They intersect in a straight line.
The angle between the two planes is defined as
the acute angle between their normal vectors.
№80 слайд
Содержание слайда: EQUATIONS OF PLANES
Find the angle between the planes
x + y + z = 1 and x – 2y + 3z = 1
Find symmetric equations for the line of intersection L of these two planes.
№81 слайд
Содержание слайда: EQUATIONS OF PLANES
The normal vectors of these planes
are:
n1 = <1, 1, 1> n2 = <1, –2, 3>
№82 слайд
Содержание слайда: EQUATIONS OF PLANES
So, if θ is the angle between the planes,
Corollary 6 in Section 12.3 gives:
№83 слайд
Содержание слайда: EQUATIONS OF PLANES
We first need to find a point on L.
For instance, we can find the point where
the line intersects the xy-plane by setting z = 0
in the equations of both planes.
This gives the equations
x + y = 1 and x – 2y = 1
whose solution is x = 1, y = 0.
So, the point (1, 0, 0) lies on L.
№84 слайд
Содержание слайда: EQUATIONS OF PLANES
As L lies in both planes, it is perpendicular to both the normal vectors.
Thus, a vector v parallel to L is given by
the cross product
№85 слайд
Содержание слайда: EQUATIONS OF PLANES
So, the symmetric equations of L can be written as:
№86 слайд
Содержание слайда: NOTE
A linear equation in x, y, and z represents
a plane.
Also, two nonparallel planes intersect in a line.
It follows that two linear equations
can represent a line.
№87 слайд
Содержание слайда: NOTE
The points (x, y, z) that satisfy both
a1x + b1y + c1z + d1 = 0
and a2x + b2y + c2z + d2 = 0
lie on both of these planes.
So, the pair of linear equations represents
the line of intersection of the planes (if they
are not parallel).
№88 слайд
Содержание слайда: NOTE
For instance, in Example 7, the line L
was given as the line of intersection of
the planes
x + y + z = 1 and x – 2y + 3z = 1
№89 слайд
Содержание слайда: NOTE
The symmetric equations that we found for L could be written as:
This is again a pair of linear equations.
№90 слайд
Содержание слайда: NOTE
They exhibit L as the line of intersection
of the planes
(x – 1)/5 = y/(–2) and y/(–2) = z/(–3)
№91 слайд
Содержание слайда: NOTE
In general, when we write the equations
of a line in the symmetric form
we can regard the line as the line
of intersection of the two planes
№92 слайд
Содержание слайда: EQUATIONS OF PLANES
Find a formula for the distance D
from a point P1(x1, y1, z1) to the plane
ax + by + cz + d = 0.
№93 слайд
Содержание слайда: EQUATIONS OF PLANES
Let P0(x0, y0, z0) be any point in the plane.
Let b be the vector corresponding to .
Then,
b = <x1 – x0, y1 – y0, z1 – z0>
№94 слайд
Содержание слайда: EQUATIONS OF PLANES
You can see that the distance D from P1
to the plane is equal to the absolute value
of the scalar projection of b onto the normal vector n = <a, b, c>.
№95 слайд
Содержание слайда: EQUATIONS OF PLANES
Thus,
№96 слайд
Содержание слайда: EQUATIONS OF PLANES
Since P0 lies in the plane, its coordinates satisfy the equation of the plane.
Thus, we have ax0 + by0 + cz0 + d = 0.
№97 слайд
Содержание слайда: EQUATIONS OF PLANES
Hence, the formula for D can be written
as:
№98 слайд
Содержание слайда: EQUATIONS OF PLANES
Find the distance between the parallel planes
10x + 2y – 2z = 5 and 5x + y – z = 1
№99 слайд
Содержание слайда: EQUATIONS OF PLANES
First, we note that the planes are parallel because their normal vectors
<10, 2, –2> and <5, 1, –1>
are parallel.
№100 слайд
Содержание слайда: EQUATIONS OF PLANES
To find the distance D between the planes,
we choose any point on one plane and calculate its distance to the other plane.
In particular, if we put y = z =0 in the equation
of the first plane, we get 10x = 5.
So, (½, 0, 0) is a point in this plane.
№101 слайд
Содержание слайда: EQUATIONS OF PLANES
By Formula 9, the distance between (½, 0, 0) and the plane 5x + y – z – 1 = 0 is:
So, the distance between the planes is .
№102 слайд
Содержание слайда: EQUATIONS OF PLANES
In Example 3, we showed that the lines
L1: x = 1 + t y = –2 + 3t z = 4 – t
L2: x = 2s y = 3 + s z = –3 + 4s
are skew.
Find the distance between them.
№103 слайд
Содержание слайда: EQUATIONS OF PLANES
Since the two lines L1 and L2 are skew,
they can be viewed as lying on two parallel planes P1 and P2.
The distance between L1 and L2 is the same
as the distance between P1 and P2.
This can be computed as in Example 9.
№104 слайд
Содержание слайда: EQUATIONS OF PLANES
The common normal vector to both planes must be orthogonal to both
v1 = <1, 3, –1> (direction of L1)
v2 = <2, 1, 4> (direction of L2)
№105 слайд
Содержание слайда: EQUATIONS OF PLANES
So, a normal vector is:
№106 слайд
Содержание слайда: EQUATIONS OF PLANES
If we put s = 0 in the equations of L2,
we get the point (0, 3, –3) on L2.
So, an equation for P2 is:
13(x – 0) – 6(y – 3) – 5(z + 3) = 0
or
13x – 6y – 5z + 3 = 0
№107 слайд
Содержание слайда: EQUATIONS OF PLANES
If we now set t = 0 in the equations
for L1, we get the point (1, –2, 4)
on P1.
№108 слайд
Содержание слайда: EQUATIONS OF PLANES
So, the distance between L1 and L2
is the same as the distance from (1, –2, 4) to 13x – 6y – 5z + 3 = 0.
№109 слайд
Содержание слайда: EQUATIONS OF PLANES
By Formula 9, this distance is:
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