Download 11.1 Three-Dimensional Coordinate System

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11.1 Three-Dimensional Coordinate System
In three dimensions, a point has three coordinates: (x, y, z). The normal orientation of the x, y, and z-axes
is shown below. The three axes divide the region into 8 octants.
z = 0 is the equation of the xy-plane.
y = 0 is the equation of the xz-plane.
x = 0 is the equation of the yz-plane.
What is the projection of the point (2, 6, −5) onto the xy-plane? the yz-plane? the xz-plane?
What do the following represent in R3 ?
y=7
2x+z = 8
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x2 + y 2 = 9
z ≥ y2
The distance between two points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is
The midpoint of the points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) is
q
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 .
x1 + x2 y1 + y2 z1 + z2
,
,
.
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2
2
The equation of a sphere with center (h, k, l) and radius r is:
(x − h)2 + (y − k)2 + (z − l)2 = r 2
Find an equation of the sphere that has center (3, −2, 7) and touches the xz-plane at a single point.
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Find the equation of the sphere that has diameter passing through the points (1, 4, −10) and (−3, 6, −2).
What is the intersection of this sphere with the yz-plane?
Find the center and radius of the sphere x2 + y 2 + z 2 + 6x − 8z = 11
Describe mathematically the top half of a solid sphere of radius 4 centered at the origin.
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11.2 Vectors and the Dot Product
In 3 dimensions a vector has 3 components: a = ha1 , a2 , a3 i. where a1 is the x-component, a2 is the
y-component, and a3 is the z-component.
An equivalent way of writing a vector is by using the standard unit basis vectors:
i = h1, 0, 0i , j = h0, 1, 0i, and k = h0, 0, 1i.
The vector a = ha1 , a2 , a3 i can be written as a = a1 i + a2 j + a3 k.
The magnitude (or length) of the vector a = ha1 , a2 , a3 i is |a| =
q
a21 + a22 + a23 .
−
−→
Given the points A(x1 , y1 , z1 ) and B(x2 , y2 , z2 ), the vector from A to B is AB = hx2 − x1 , y2 − y1 , z2 − z1 i.
a
To find a unit vector (vector of length 1) in the direction of a vector a, compute
.
|a|
Let a be the vector from the point P (2, −4, −7) to the point Q(1, 3, −5) and let b = −4i + 2j − 6k.
Fid a unit vector in the same direction as a.
Find a vector of length 4 in the same direction as the vector a + 2b.
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Two vectors are parallel if one vector is a scalar multiple of the other.
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D
For example, a = h4, −3, 6i is parallel to b = − 34 , 1, −2 since a = −3b.
Given two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i, the dot product (or scalar product) of a and b,
denoted a · b, can be found in either of the following ways:
a · b = |a||b| cos θ
where θ is the angle between a and b
a · b = a1 b1 + a2 b2 + a3 b3
A dot product can only be performed on two vectors and the result is a scalar.
Note: In the context of this section, a · b would make sense, but |a| · b would not since |a| is not a vector.
The · here does not mean multiplication. It means dot product.
The first formula above rearranged gives a formula for finding the cosine of the angle between two vectors.
cos θ =
a·b
|a||b|
Two vectors a and b are orthogonal (or perpendicular), if a · b = 0.
For what values of x are the vectors hx, 3x, 4i and hx, 4, 5i orthogonal?
A triangle has vertices A(0, 3, 9), B(1, −2, 1), and C(3, 1, 2). Find 6 ABC.
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Given two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i,
The scalar projection of b onto a is given by
compa b =
a·b
|a|
The vector projection of b onto a is given by
proja b =
a·b
|a|
a·b
a
=
a
|a|
|a|2
Given the vectors a = 10i − 2k and b = h3, −4, 1i, find the scalar and vector projections of b onto a.
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11.3 The Cross Product
A determinant for a 2 × 2 array of numbers (matrix):
= ad − bc
2 −3
Example: 5 −9
a b
c d
=
A determinant for a 3 × 3 array of numbers (matrix):
a
1
b1
c1
a2 a3 b b b b b b 2
1
2 1
3
3
b2 b3 = a1 + a3 − a2 c1 c2 c1 c3 c2 c3 c2 c3 2
Example: 0
5
3 1 2 1 =
3 0 The cross product of two vectors a = ha1 , a2 , a3 i and b = hb1 , b2 , b3 i is:
i
a × b = a1
b1
Example: Find a × b if a = h3, 2, −1i and b = h4, 1, 1i.
The cross product of two vectors is a VECTOR!
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j k a2 a3 b2 b3 c
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Very Important Fact: The vector a × b is orthogonal to both a and b.
Find a vector that is perpendicular to the plane containing the points A(1, 2, 3), B(−2, 1, −1), and C(1, −1, 1).
The direction in which the cross product points can be determined by the right-hand rule.
The right-hand rule helps us to see that a × b 6= b × a. What is true is a × b = −b × a.
If θ is the angle between two vectors a and b, then
|a × b| = |a||b| sin θ
This above fact tells us the following:
(1) Two nonzero vectors a and b are parallel if and only if |a × b| = 0.
(2) The area of the parallelogram formed by the vectors a and b is |a × b|.
Find the area of the triangle from the previous example with vertices A(1, 2, 3), B(−2, 1, −1), and C(1, −1, 1).
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The scalar triple product of the vectors a, b, and c is a · (b × c).
The volume of the parallelipiped determined by the vectors a, b, and c is the absolute value of the scalar
triple product:
V = |a · (b × c)|
When finding the scalar triple product, you can either first find b × c and then dot with a, or you can
find it all in one step by computing the determinant below where a =< a1 , a2 , a3 >, b =< b1 , b2 , b3 >, and
c =< c1 , c2 , c3 >.
a
1
b1
c1
a2 a3 b2 b3 c2 c3 Find the volume of the parallelipiped formed by the vectors a = h1, 3, 1i, b = h4, −1, 2i and
c = h2, 2, 0i.
What does it mean, then, if the scalar triple product of three vectors is 0?
Do the points P (3, 0, 1), Q(−1, 2, 5), R(5, 1, −1) and S(0, 4, 2) all lie in the same plane? i.e. Are they
coplanar?
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