Electrodynamics

To find a unit vector that has the same direction as vector v... Ex: Find the unit vector in the same direction as v = 5i - 12j Then verify that the magnitude of this new unit vector is 1

...divide the vector v by it's magnitude (5i - 12j)/√(5^2+12^2) = (5i - 12j)/13 = 5/13i - 12/13 j Verify: (5/13)^2 + (-12/13)^2 = √(169/169) = 1

v = 4i + j w = 2i - j Find v dot w

1st equation for dot product, using coordinates: v dot w = (4*2) + (1*-1) = 7 2nd equation for dot product cos(theta) is better explained in picture but simply: v dot w = |v| * |w| * cos(theta) = √17 * √5 * (projvw/√5) = √17 * projvw = √17 * 1.69 = 7

The magnitude and direction exerted by two tugboats towing a ship are 1610 kilograms, N35W, and 1250 kilograms, S55W, respectively. ind the magnitude, to the nearest kilogram, and the direction angle, to the nearest tenth of a degree, of the resultant force.

2038.28 lb; 162.8 degrees add up the i's and j's to get the new vector, expressed in i and j's. Then do pythag theorem to get the magnitude of the vector. Then do inverse cosine to get degrees

Add the two vectors: 4 units east and 3 units north.

5 units at an angle of 36.87 from the positive x axis And it is pointing away from the origin of course

Scalar multiplication of a vector written in terms of i and j ex: v = 5i + 4j Find: 6v Find: -3v

6v = 30i + 24j -3v = -15i - 12j

v = 5i + 4j w = 6i - 9j Find 4v - 2w

8i + 34j

Vector triple product: What is the BAC-CAB rule?

A x (C x B) = B(A*C) - C(A*B)

Equation for cross product of two vectors (cross product also called vector product)

A x B = A * B sin(theta) * n^ n^ (n with a hat on it) is just a unit vector pointing perpendicular to x,y plane The cross product takes two vectors, puts them tail to tail, then makes a parallelogram out of them and solves for its area. The area that is found is then used as the magnitude for a vector pointing perpendicular to the x, y plane (either into the page or out). This new vector is the end product of cross product. We use absin(theta) rather than simply base * height because we don't know the height and b*sin(theta) solves for the height. Other method of finding cross product in picture (determinant)

To subtract a vector...

Add its opposite

Vector v = ai + bj can be written terms of the magnitude of v and the angle theta Do example: Wind is blowing at 20 miles per hour in the direction N30W (30 degrees left of y axis). Express its velocity as a vector v in terms of i and j

Angle theta is the angle that v makes with the positive x-axis (same as unit circle). This angle is called the "direction angle" of v. So if you draw out a vector, and then a and b (its legs), by definition: cos(theta) = a/||v|| sin(theta) = b/||v|| Therefore: v = ai + bj can also be written as: v = (||v|| * cos(theta) * i) + (||v|| *sin(theta) * j) Example: v = 20 cos(120) * i + 20 sin(120)j = 20(-1/2)i + 20(√3/2)j Therefore v = -10i + 10√3j

Properties of cross product

Anti-commutative: i x j = -j x i = k(hat) j x k = -k x j = i(hat) k x i = -i x k = j(hat) Sin(theta): (cuz sin(0) = 0) i x i = 0 j x j = 0

Prove the BAC-CAB rule, that A x (B x C) = B(A*C) - C(A*B) (just for x hat, the rest follows due to symmetry)

Cross product of B and C is: x-hat[Ay(BxCy − ByCx) − Az(BzCx - BxCz)] then cross product of A and (BxC): = x-hat(AyBxCy - AyByCz - AzBzCz + AzBxCz) B(A*C) - C(A*B) = [Bx(AxCx + AyCy + AzCz) - Cx(AxBx + AyBy + AzBz)x-bar = x-bar(AyBxCy + AzBxCz - AyByCx - AzBzCx)

Prove that [A × (B × C)] + [B × (C × A)] + [C × (A × B)] = 0

Do BAC-CAB for each and they minus each other

Are these vectors orthogonal?: v = 6i - 3j w = i+2j

Dot product (using the equation without the cos(theta) is: v*w = (6*1) + (-3*2) = 0 Because the vectors are nonzero, the zero dot product must result from theta being 90, or cos(theta) = 0 (because the other equation is ||v|| * ||w|| * cos(theta) so if v and w are not 0, cos(theta) must be 0)

Two forces, F1 and F2, act on an object. The magnitude and direction of F1 is 10 pounds and N20E. F2 is 30 pounds and N65E. Find the magnitude and the direction of the resultant force, expresses to the nearest hundredth of a pound and tenth of a degree.

F1 = 10 cos(70) i + 10sin(70)j = 3.42i + 9.40j F2 = 30 cos(25) i + 30 sin(25) j = 27.19i + 12.68j F = F1 + F2 = (3.42 + 27.19)i + (9.4 + 12.68)j = 30.61i + 22.08j Now to find its magnitude: ||F|| = √(30.61^2 + 22.08^2) = 37.74 Its angle is: cos(theta) = 30.61/37.74 theta = cos^-1(30.61/37.74) = 35.8 degrees All in all: Magnitude of 37.74, and an angle of 35.8

A vector has direction and magnitude but what does it not have?

Location. A displacement of 4 miles north from Boca is represented with same vector as is a displacement of 4 miles north from Miami

A vector multiplied by a scalar...

Multiplies the magnitude of the vector but leaves the direction unchanged. Except if the scalar is negative, the vector's direction reverse negative it's reversed

Is A x (B x C) = C x (A x B)?

No; A x (B x C) = -C x (A x B) Cross products are not associative

Find the separation vector from the source point (2,8,7) to the field point (4,6,8). Determine its magnitude, and construct the unit vector.

Seperation vector = 2i - 2j + k Unit vector: = 2/3i - 2/3j + 1/3k

Use the cross product to find the components of the unit vector nˆ perpendicular to the shaded plane in Fig. 1.11. (look at answer real quick to see the figure

Solutions just shows use cross product: = 6i+3j+2k But because we need a unit vector (of magnitude 1) (6/7)i + (3/7)j + (2/7)k

What does orthogonal mean, and what does the dot product of two orthogonal vectors equal?

You use the word orthogonal instead of perpendicular when talking about vectors that meet at right angles. ||v|| * ||w|| * cos(90) = ||v|| * ||w|| * 0 Therefore = 0

a) Can you make a cross product from a scalar and a vector? Ex: A x (B*C) b) How can a scalar triple product be reorganized and still be equivalent?

a) No b) You can move the A B and C around as long as they're in the same order Ex: A*(BxC) = C*(AxB) = B*(CxA)

Scalar triple product A * ( B x C) a) What is geometry of it? b) How to solve it with matrix?

a) So B x C gives you area of a parallelogram and uses this as the magnitude of the vector that is perpendicular to it. Then A * this new perpendicular vector gives you the projection of A onto that perpendicular vector, multiplied by the perpendicular vector itself. So essentially it's base times height times altitude. b) Take determinant of matrix with A on top (A, or whichever vector is the dot product one, is basically the i j k)

Find the angle between two face diagonals of a cube (look at pic for visual)

cos(theta) = (v * w)/(||v||*||w||) so cos(θ) = ((0*1) + (1*0) + (1*1))/(√2 * √2) cos(θ) = 1/2 so θ = 60 degrees Could have gotten answer more easily by drawing in a diagonal across the top of the cube, completing the equilateral triangle.

Find the angle between two body diagonals of a cube

cos(θ) = (v*w)/(||v||*||w||) so cos(θ) = ((1*1) + (1*1) + (1*-1))/((√3 * √3) cos(θ) = 1/3 θ = 70.5288

v = 2i + 4j w = -2i + 6j Find the vector projection of v onto w

proj(w)v = v*2/(||w||^2) * w (2i + 4j) * (-2i + 6j)/√(-2^2 + 6^2) * w = 20/40 * w = 20/40 * (-2i + 6j) = -i + 3j

The 2 equations for dot product of two vectors v and w (dot product is also called scalar product) Equation for angle between two vectors

v * w = (a1 * a2) + (b1 * b2) v * w = ||v|| * ||w|| * cos(theta) where theta is the angle between the two vectors when placed tail-to-tail The dot product of two vectors is a scalar cos(theta) = (v * w)/(||v||*||w||)

To add two vectors that are written in i,j form, just line it up and add Ex: vector v = 5i + 4j vector w = 6i-9j What is v+w? What is v - w

v + w = (5+6)i + (4-9j) = 11i - 5j v - w = (5-6)i + (4+9)j = -i + 13j

To put a vector in standard form, starting at the origin, express it in terms of unit vectors i and j. Ex: Vector, v, begins an initial point P1 = (3,-1) and goes to terminal point P2 = (-2,5). Express v as starting at the origin by writing v in terms of i and j.

v = (x2-x1)i + (y2-y1)j v = (-2-3)i + (5-(-1)j = -5i + 6j

Definition for magnitude of vector that is a product of 3 vectors (3d plane)

| A |→ = √(Ax^2 + Ay^2 + Az^2) (pythag theorem works even in 3d)

What is determinant formula for 3x3

|a b| (ad-bc) - (ad-bc) + (ad-bc) |c d|

Two equations for projection of a vector (v) onto another vector (w) The projection line is going to be wherever the end point of v is perpendicular to the ghost of w

||proj(subscriptw)v|| = ||v|| * cos(theta) - where theta is the angle between v and w equation 2: proj(subw)v = (v * w)/(||w||^2) and then all of that times w