Vector Analysis and Quaternions
Product of Two Vectors
Excerpts
Product of Two Vectors
The formula $\text{velocity flux} = \text{electromotive-force}$ is much handier than any thumb-and-finger rule; for it compares the three directions directly with the right-handed screw.
Product of Two Vectors
The square combinations give results which are independent of direction, and consequently are summed by simple addition.
- This excerpt is in Products of Coplanar Vectors (Products of Coplanar Vectors)
Product of Two Vectors
The vector product as before is denoted by $\mathrm{V}AB$. It means the product of $A$ and the component of $B$ which is perpendicular to $A$, and is represented by the area of the parallelogram formed by $A$ and $B$.
Product of Two Vectors
The product is positive when the vector and the projection have the same direction, and negative when they have opposite directions.
Product of Two Vectors
Frequently all that is demanded is, given two of these directions to determine the third.
Product of Two Vectors
In a sum of vectors, the vectors are necessarily homogeneous, but in a product the vectors may be heterogeneous.
Equations
Product of Two Vectors
ij &= k, & jk &= i, & ki &= jThe right-handed cyclic rules for products of the unit vectors in space: i times j is k, j times k is i, and k times i is j, taken as assumed by symmetry of space.
Product of Two Vectors
\text{velocity flux} = \text{electromotive-force}In the dynamo, the velocity of the conductor combined with the magnetic flux with the right-handed screw rule gives the direction of the electromotive force. The chapter states it in words, so no letter symbols are defined.
Product of Two Vectors
\text{current flux} = \text{mechanical-force}In the electric motor, the current in the conductor combined with the magnetic flux gives the direction of the mechanical force on the conductor. The chapter states it in words, so no letter symbols are defined.
Product of Two Vectors
\text{flux force} = \text{current}Cyclical permutation of the motor formula gives flux and force producing current, stated in words.
Product of Two Vectors
A &= a_1j + a_2j + a_3kDefines the vector A by its three components along the axes. The chapter prints a_1j for the first term where a_1i is evidently meant; this is a typesetting slip in the source, flagged here and not corrected.
Product of Two Vectors
a_1b_1 + a_2b_2 + a_3b_3 =\mathrm{S}ABThe scalar product of A and B is the sum of the products of their corresponding components; it is independent of direction.
- This equation is in Products of Coplanar Vectors (Products of Coplanar Vectors)
Product of Two Vectors
A^2 = {a_1}^2 + {a_2}^2 + {a_3}^2 = a^2The square of a vector is the sum of the squares of its components, which equals the square of its magnitude a, and is always positive.
- This equation is in Products of Coplanar Vectors (Products of Coplanar Vectors)
- This equation is in Products of Coplanar Vectors (Products of Coplanar Vectors)
- This equation is in Products of Coplanar Vectors (Products of Coplanar Vectors)
Product of Two Vectors
\mathrm{S}SB = b_ls_l + b_2s_2 + b_3s_3The magnetic flux through an area S is the scalar product of the flux intensity B with the area vector S. Flagged: the chapter prints b_l s_l, presumably b_1 s_1, and the operator as SSB, read here as S applied to S and B.
Product of Two Vectors
(A + B)C = AC + BCThe vector product distributes over a sum of vectors in the first factor: the product of the sum A+B with C equals AC plus BC.
Product of Two Vectors
(A+B)(C+D) = AC + AD + BC + BDThe product of two sums of non-successive vectors expands term by term, each sum distributing over the other.
Product of Two Vectors
(A+B)^2 & = A^2 + B^2 + AB + BAThe square of a sum of vectors expands to the squares of each vector plus the product of A and B and the product of B and A.
Product of Two Vectors
A^2 + B^2 + 2\mathrm{S}ABFor a sum of vectors, the square equals the sum of the squares of the two vectors plus twice their scalar product.
Problems
Exercise Probs-28-37
Exercise Probs-28-37, problem 28
The relative velocity of a conductor is S.W., and the magnetic flux is N.W.; what is the direction of the electromotive force in the conductor?
Printed answer:- (none printed)
unverified: no computed check settled this one (yet)
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other: not a kind the checker handles
Exercise Probs-28-37, problem 29
The direction of the current is vertically downward, that of the magnetic flux is West; find the direction of the mechanical force on the conductor.
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Exercise Probs-28-37, problem 30
A body to which a force of $2i + 3j + 4k$ pounds is applied moves with a velocity of $5i + 6j + 7k$ feet per second; find the rate at which work is done.
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Exercise Probs-28-37, problem 31
A conductor $8i + 9j + 10k$ inches long is subject to an electromotive force of $11 i +12j + 13k$ volts per inch; find the difference of potential at the ends.
Printed answer:- 326 volts.
verified: the printed answer passed a computed check
How it was checked
evaluate: passes326
On the STU-32 (STU, rpn):
8 ENTER 11 ×
9 ENTER 12 × +
10 ENTER 13 × +
Calculator:
+326E+0; the book prints326. Run on the calculator core at firmware628c96c.Exercise Probs-28-37, problem 32
Find the rectangular projections of the area of the parallelogram defined by the vectors $A = 12i - 23j - 34k$ and $B = -45i - 56j + 67k$.
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Exercise Probs-28-37, problem 33
Show that the moment of the velocity of a body with respect to a point is equal to the sum of the moments of its component velocities with respect to the same point.
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Exercise Probs-28-37, problem 34
The arm is $9i + 11j + 13k$ feet, and the force applied at either end is $17i + 19j + 23k$ pounds weight; find the torque.
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Exercise Probs-28-37, problem 35
A body of 1000 pounds mass has linear velocities of 50 feet per second $\overline{30^\circ/}\!\underline{/45^\circ}$ and 60 feet per second $\overline{60^\circ/}\!\underline{/22^\circ.5}$; find its kinetic energy.
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Exercise Probs-28-37, problem 36
Show that if a system of area-vectors can be represented by the faces of a polyhedron, their resultant vanishes.
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Exercise Probs-28-37, problem 37
Show that work done by the resultant velocity is equal to the sum of the works done by its components.
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