代做21-259: Calculus in Three Dimensions Lecture #2 Spring 2025调试SPSS
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Lecture #2
Spring 2025
The Cross Product
Definition: A 2×2 determinant of the matrix is the quantity
For a 3×3 matrix, we can compute the determinant as follows:
Example 1. Compute the determinants and
Definition: Let u = 〈u1,u2,u3〉 and v = 〈v1, v2, v3〉 be vectors in R 3 . The cross product (or vector product) of u and v is the vector given by u × v = 〈u2v3 −u3v2,u3v1 −u1v3,u1v2 −u2v1〉.
The cross product of u and v is really a determinant:
Note: the cross product is defined only for vectors in R3 - there is an analogous quantity for vectors in R 2 (although it is interpreted as a scalar) and it can be generalized to higher dimensions, but in our context a cross product will only be computed for vectors in R3.
u × v = u2v3ı +u3v1 ȷ +u1v2k −u2v1k −u3v2ı −u1v3 ȷ
Example 2. Compute the following cross products:
a) (ı + ȷ +k)×(ı −2ȷ +2k)
b) 〈−1,−3,1〉 × 〈1,2,−2〉
Properties of the Cross Product: Let u, v, and w be vectors in R 3 and let a ∈ R. Then we have:
1. u × v = −v ×u (anticommutativity of ×)
2. (au)× v = a(u × v) = u ×(av) (associativity of scalar multiplication)
3. u ×(v +w) = u × v +u ×w (left distribution of × across + )
4. (u + v)×w = u ×w + v ×w (right distribution of × across + )
Also, by using vectors of zeros and ones, it is easy to show the following:
ı × ȷ = k ȷ ×k = ı k × ı = ȷ
ȷ × ı = −k k × ȷ = −ı ı ×k = −ȷ
ı × ı = 0 ȷ × ȷ = 0 k ×k = 0
Geometrically, the cross product of two vectors u and v results in a vector u × v that is orthogonal to both u and v. The direction of the vector is given by the right-hand rule - as the fingers of your right hand curl from u to v, your thumb points in the direction of u × v. The magnitude of the cross product vector is
|u × v| = |u||v|sinθ
where 0 ≤ θ ≤ π.
If u and v are parallel, then u × v = 0. If u and v are orthogonal, then |u × v| = |u||v|.
The cross product can then be used to compute a vector that is orthogonal to two given vectors.
Example 3. Find a unit vector that is orthogonal to both u = 〈2,3,−5〉 and v = 〈−1,1,−1〉.
Example 4. Find the magnitude of u × v if u = 〈1,−1,3〉 and v = 〈−2,0,−1〉.
Example: What is the relationship between u · v and |u × v|?
Example: If u · v = 0 and u and v are nonzero vectors, is it possible for u × v = 0?
There are several geometric applications of the cross product. First, the magnitude of u × v gives the area of the parallelogram formed by u and v.
Area = |v||u|sinθ
Example 5. Find the area of the parallelogram formed by the vectors u = 〈1,0,1〉 and v = 〈2,4,2〉.
Definition: The scalar triple product of the vectors u,v, and w is the scalar quantity u ·(v ×w).
The absolute value of the scalar triple product gives the volume of a parallelepiped formed by the vectors u, v, and w.
Note that it does not matter what order the three vectors are written when forming the parallelepiped:
|u ·(v ×w)| = |v ·(u ×w)| = |w ·(u × v)|
The tetrahedron formed by the vectors u, v, and w has volume 1/6 of the volume of the corresponding parallelepiped.
Definition: The vector triple product of the vectors u, v, and w is the vector u ×(v ×w).
Example 6. Let u, v, and w be vectors in R 3 . Which of the following expressions are meaningful? If not, explain why, and if so, state whether the result is a vector or scalar.
a) (u · v)×w
b) (u ×w)·(v ×w)
c) u ·(v ·w)
d) u ·(v ×w)
Properties of the Vector Triple Product: Let u, v, and w be vectors in R3 and let a ∈ R. Then we have:
1. u ×(v ×w) = (u ·w)v −(u · v)w
2. (u × v)×w = −w ×(u × v) = −(w · v)u +(w ·u)v
3. u ×(v ×w)+ v ×(w ×u)+w ×(u × v) = 0
4. (u × v)×w = u ×(v ×w)− v ×(u ×w)
Other useful identities:
— (u × v)·(u × v) = |u × v|2 = |u|2 |v|2 −(u · v)2
— (t ×u)·(v ×w) = (t · v)(u ·w)−(t ·w)(u · v)
Torque, or moment of force, is the tendency of a force to rotate an object about an axis. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Consider a rigid body situated at the origin of a coordinate system. If a force F acts on a rigid body at a point whose position is given by r , then the rotational force exerted on the body is the torque τ given by τ = r ×F. The torque vector has a magnitude of
|τ| = |r ×F| = |r ||F|sinθ.
Since |F| has units of Newtons (N), or kg-m/s2 , the magnitude of torque has units of Newton-meters (N-m).
Lines in Space
Lines in R2: We have known for some time that a line in R2 is determined by a point on the line and the slope of the line. Let a represent the change in x and b represent the corresponding change in y of the slope (if the line is vertical then a = 0). The slope m = b/a can be interpreted as a direction vector: v = 〈a,b〉. If a ≠ 0 and the line passes through the point (x0, y0), then we know its equation is given by
which can also be written as (if b ≠ 0)
Using vectors, we can also describe the line through (x0, y0) and slope m = b/a. Now any point on the line (x, y) lies somewhere in the direction of the slope vector v (we call v the direction vector of the line). In other words, there is some scalar multiple of v such that the position vector of the point (x, y) is the sum of the scaled v and the position vector r 0 of the known point (x0, y0). Thus there is some value of a parameter t (a scalar) such that the position vector for the point (x, y) is given by
〈x, y〉 = r (t) = r 0 + tv = 〈x0, y0〉 + t〈a,b〉 = 〈x0 + at, y0 +bt〉.
In this case, the equation r (t) = r 0 + tv is known as the vector equation of the line L, and the two equations x = x0 + at, y = y0 +bt are the parametric equations of L. Note that if a = 0, then the line is vertical, so in this case the x-coordinate of any point on the line is x0. This yields the vector equation r (t) = 〈x0, y0〉 + t〈0,b〉, and thus the parametric equations are x = x0, y = y0 + bt. If b = 0, then y is always y0, and the vector equation is r (t) = 〈x0 + at, y0〉, and the parametric equations are x = x0 + at, y = y0.
Definition: Let L represent the nonvertical line in R 2 passing through (x0, y0) with slope m = b/a. Then the following equations represent all points on L:
Symmetric Equations: a/x − x0 = b/y − y0
Vector Equation: r (t) = r 0 + tv = 〈x0, y0〉 + t〈a,b〉
Parametric Equations: x = x0 + at, y = y0 +bt
Example 7. Find the symmetric, vector, and parametric equations of the line passing through the points (−2,3) and (4,1). Then find an alternate parametrization of the line (i.e., another set of parametric equations that describe the same line).
Example: How many different parametrizations does a line have?
Two lines in R 2 are either parallel (i.e., have the same slope) or they intersect at a single point. The point of intersection (x, y) must lie on both lines, so the coordinates of the point of intersection must satisfy the equations of both lines. However, if using the parametric or vector equations of the lines, note that the values of t at the point of intersection may be different for each. Thus,if using parametric equations or vector equations to solve for a point of intersection, then you may want to use a different variable (e.g. s) for the parameter of one of the lines.
Example 8. Find the coordinates of the point of intersection of the lines L1 and L2 given by
L1 : r 1(t) = 〈3− t,2+ t〉, and L2 : r 2(s) = 〈4+2s,−3s〉.
Lines in R 3 : Lines in three spatial dimensions are described in the same way as they are in R2 . One major difference however is that the concept of slope does not readily extend to R 3 - the change in y with respect to x is no longer sufficient to describe the direction of the line. Thus in order to describe a line, we must have a point (x0, y0, z0) on the line and a vector v = 〈a,b,c〉 that gives the direction. The numbers a, b, and c are known as the direction numbers of the line.
Symmetric Equations: a/x − x0 = b/y − y0 = c/z − z0
Vector Equation: r (t) = r 0 + tv = 〈x0, y0, z0〉 + t〈a,b,c〉
Parametric Equations: x = x0 + at, y = y0 +bt, z = z0 +c t
Note that if one of the direction numbers is 0 the symmetric equations will be slightly different. For example, if a = 0, then L would be represented by the symmetric equations
x = x0, b/y − y0 = c/z − z0.
Example 9. Find the symmetric, vector, and parametric equations of the line through the points (6,1,−3) and (2,4,5).
While it is certainly the case that parallel lines do not intersect, in R 3 there are nonparallel lines that do not intersect.
Definition: Two lines L1 and L2 in R 3 are skew if they are not parallel and do not intersect.
Example 10. Determine whether the following lines are parallel, skew, or intersecting. If they intersect, find the point of intersection.
a) L1 : x = −6t, y = 1+9t, z = −3t, L2 : x = 1+2s, y = 4−3s, z = s
b) L1 : 1/x = 2/y −1 = 3/z −2, L2 : −4/x −3 = −3/y −2 = 2/z −1
c) L1 : r (t) = 〈1− t,2+ t,−2−4t〉, L2 : r (s) = 〈−2s,2s +3,−8s −6〉
To parametrize a line segment between two points (x0, y0, z0) and (x1, y1, z1), one can simply take a convex combination of the position vectors r 0 and r 1 of the two points:
r (t) = (1− t)r 0 + tr 1, 0 ≤ t ≤ 1.
This just amounts to taking a weighted average of the two position vectors, with the extremes being r 0 when t = 0 and r 1 when t = 1.