Two vectors v and w are parallel if there exists a constant number c such that v = c * w. If you are given the components of v and w this condition can be written out as a vector equation. This equation is satisfied if and only if each component of the left-hand side is equal to the corresponding component on the right.
If we are given three points, say P, Q and R, in the plane, then they form a triangle whose sides are the magnitudes of the vectors PQ, PR and QR. If the coordinates of P, Q and R are given, then we can write out the three vectors in terms of these coordinates, and we can then write down the magnitudes of the three. The triangle will be a right triangle if and only if the Pythagorean Theorem is satisfied, i.e., if and only if the square of the longest side is equal to the sum of the squares of the two shorter sides. An independent test for a right triangle: If the dot product of two of the vectors is zero, the angle between them is 90 deg and we can again conclude that the triangle is a right triangle. The triangle is isosceles if two of the sides are equal, i.e., if the magnitudes of two of the vectors are equal. It is possible for a triangle to be both right and isosceles, in which case the angles of the triangle would be 45 deg, 45 deg and 90 deg.
The test for orthogonality of two vectors is that their dot product be zero. The dot product will be a single number or algebraic expression. Setting this expression equal to zero will give us an equation which expresses the orthogonality of the two vectors.
To find a vector orthogonal to two given vectors, it's easy to use the cross product. Dividing the cross product by its magnitude we get a unit vector orthogonal to the two given vectors. The negative of this vector is also orthogonal to the two given vectors. Given three nonzero vectors, none parallel to any of the others, we can use this reasoning to find a vector perpendicular to the first two. If the third is also perpendicular to this vector, then it lies in the same plane as the first two.
To find a vector orthogonal to two given vectors, we can also use the dot product. Let the two given vectors be u and v, and let desired vector be u = <a, b, c> , where a, b and c are parameters to be evaluated. The condition of orthogonality tells us that u dot w = 0, and u dot v = 0. This gives us two equations in the three unknowns a, b and c. For example if v = <1, 2, -2> and w = <1, 1, -2> , then we get the two equations
a + 2 b - 2 c = 0 and
a + b - 2 c = 0.
These equations can be solved for any two of the parameters a, b and c, in terms of the third.
All vectors in a plane are perpendicular to any vector perpendicular to the plane. To determine whether 3 vectors, none of which are parallel to any of the others, are all in the same plane, first find a vector, which we will call N, perpendicular to two of the three. Then figure out whether the third vector is also perpendicular to N. To find a vector perpendicular to the first two find their cross product.
You have probably submitted your access code request for physics.
You should go ahead and submit one also for vector calculus.
If you have questions between classes, you can submit them using the Submit Question Form at http://vhcc2.vhcc.edu/dsmith/forms/question_form.htm .
At the site http://vhcc2.vhcc.edu/pc2fall9/frames%20pages/class_notes.htm you will find lecture notes for my Precalculus II class. The links listed below address trigonometric functions and on conic sections, both topics that will be relevant to your course in the next week or two. At times I might refer you to these notes to amplify certain topics.
At this point you should spend an hour or so just skimming over these notes, so you have an overview of what's there. Any video links within these notes won't work, but I can give you a disk that contains the same notes, along with the videos. On the disks, those links will work.
You should then be review in some detail the circular model of trig functions (#01, #02 and #03).
#01: First Circular Model; Introduction to Radian Measure
#02: Sketching Exercises
#03: Modeling Periodic Phenomena with Circles
#04: The Sine Function
#05: Modeling with the Sine Function
#06: Trigonometric Identities
#12: Introduction to Conic Sections
#13: Circles, Ellipses, Hyperbola
#14: Conic Sections at General Positions; Parametric Equations
#15: Parametric Equations
#16: Polar Coordinates
Equation of a circle or a sphere
In two dimensions a circle of radius 4 centered at the point (3, 2) consists of the set of all points which lie 4 units from (3, 2).
If (x, y) is a point in the plane, its distance from (3, 2) is sqrt( (x - 3)^2 + (y - 2)^2 ).
The point (x, y) will lie on the given circle if, and only if, its distance from (3, 2) is 4.
Thus the condition for a point to lie on the circle is
- sqrt( (x - 3)^2 + (y - 2)^2 ) = 4.
We can square both sides and get the equivalent equation
- (x - 3)^2 + (y - 2)^2 = 16.
In general if (h, k) is the center of a circle of radius r, the same reasoning leads us to the general equation of the circle:
The distance between (x, y) and (h, k) is sqrt( (x - h)^2 + (y - k)^2 ).
If this distance is r, then the square of this distance is r^2 and we have
- (x - h)^2 + (y - k)^2 = r^2.
This isn't a formula to be memorized, but rather a picture and a reasoning sequence to be understood.
In three dimensions everything works out very similarly. For example the sphere of radius 7 centered at (-4, 6, 12) is characterized by points (x, y, z) which lie at distance 7 from the center, so the equation is
(x - (-4)) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 7 ^ 2, which simplifies to
(x + 4) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 49.
The reasoning used in the above is exactly analogous to that used previously in two dimensions. The equation of a sphere of radius r centered at (x_0, y_0, z_0) is found by the same sequence of reasoning to be
(x - x_0) ^ 2 + (y - y_0) ^ 2 + (z - z_0)^2 = r^2
Returning to the equation (x + 4) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 49, which is in one standard form for the equation of a sphere, we put this equation into the other standard form by expanding the squares and subtracting 49 from both sides. We get
x^2 + y^2 + z^2 + 8 x - 12 y - 24 z + 147 = 0.
It's easy to move from the first standard form to the second, as we have just done. It takes a little more work to move in the other direction. Let's pretend we didn't already know the center and radius of the sphere given by the equation x^2 + y^2 + z^2 + 8 x - 12 y - 24 z + 147 = 0. How would we recover this form?
The answer is that we complete the squares on the x, y and z terms. We begin by writing the equation in the form
x^2 + 8 x + y^2 - 12 y + z^2 - 24 z = - 147.
Completing the square on x^2 + 8 x, we find that x^2 + 8 x + 16 is the perfect square of (x + 4). If we replace x^2 + 8 x by the equal quantity x^2 + 8 x + 16 - 16, we will not change the balance of the equation. We will then be able to rewrite this new term as
Similarly we will replace y^2 - 12 y by y^2 - 12 y + 36 - 36 and z^2 - 24 z by z^2 - 24 z + 144 - 144. Our equation will become
x^2 + 8 x + 16 - 16 + y^2 - 12 y + 36 - 36 + z^2 - 24 z + 144 - 144 = -147, which can be rewritten as
(x + 4) ^ 2 - 16 + (y - 6) ^ 2 - 36 + (z - 144) ^ 2 - 144 = -147. Then adding 16 + 36 + 144 to both sides we get
(x + 4) ^ 2 + (y - 6) ^ 2 + ( z - 12 ) ^ 2 = 49, which agrees with the first standard form.
Unit vectors
Three problems were given:
1. Prove that cos ( theta) `i + sin ( theta ) * `j is a unit vector.
2. Find a unit vector in the direction of the vector 2 `i + 5 `j. Then find theta such that cos(theta) i + sin(theta) j is that unit vector.
The solutions:
Problem 1:
The magnitude of cos ( theta) `i + sin ( theta ) * `j is
|| cos ( theta) `i + sin ( theta ) * `j || = sqrt( (cos(theta)) ^ 2 + (sin(theta))^2 ), also written as sqrt( cos^2(theta) + sin^2(theta) ).
By the unit-circle definition of the sine and cosine functions, cos^2(theta) + sin^2(theta) = 1 for any theta.
Thus the magnitude of cos ( theta) `i + sin ( theta ) * `j is 1, for any value of theta.
note the importance of understanding the unit-circle definition of the sine and cosine functions
Problem 2:
The magnitude of 2 `i + 5 `j is sqrt( 2^2 + 5^2) = sqrt( 29 ).
The requested unit vector is therefore ( 2 `i + 5 `j ) / sqrt(29) = 2 sqrt(29) / 29 `i + 5 sqrt(29) / 29 `j.
To 2 significant figure this is .37 `i + .93 `j.
If this is in the form cos(theta) `i + sin(theta) `j, then since two vectors are equal if an only their components are equal, we have
cos(theta) = .37 and
sin(theta) = .93.
We can sketch a unit circle, with the point (.37, .93) and a radial line to that point. We estimated that angle as about 70 or 80 degrees. Then plugging into our calculator we found that
theta = arcCos(.37) = 1.19.
We also find that
theta = arcSin(.93) = 1.19.
The two agree, which is a good thing because there is only one theta. There will be a slight difference in the results because we used two-significant-figure approximations for the components of the vector. The exact value of the angle would be arcCos(2 sqrt(29) / 29), which is exactly the same as arcSin(5 sqrt(29) / 29).
Equality of Two Vectors
Two vectors are equal if, and only if, all their components are equal.
So, for example, if A = 4 s `i + 5 z `j + (z - r) `k and B = (z + r) `i + 10 + 3 z `k we have the following:
z + r = s, since the `i components are equal
5 z = 10 since the `j components are equal
z - r = 3 z since the `k components are equal
This gives us three equations, which we can solve for the three unknowns to get z = 2, r = 4 and s = 6.
As another example where this principle is applied:
If u = - 4 `i + 3 `j and v = 2 `i - 1/2 `j, find the values of a and b such that a * 3 `j + b * u = v.
We write the equation a * 3 `j + b * u = v out as
3 a `j + b ( -4 `i + 3 `j) = 2 `i - 1/2 `j. We simplify the left-hand side to obtain
-4 b `i + (3 a + 3 b) `j = 2 `i - 1/2 `j. Setting the `i and `j components equal we get
-4 b = 2 and
3 a + 3 b = -1/2.
We easily solve this system, obtaining b = -1/2 and a = 1/3.
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Here are some exercises designed to give you a head start on some of next week's topics. You have plenty to do with the problem assignments (and I'm sure you'll have some questions on Monday), but do try to take some time to do the following. Put a 15-minute limit on yourself for each questions; they're worth that much time, but don't let them bog you down.
1. If u = 3 `i + 5 `j , and we know that u dot v = 0 for the vector v = a `i + 4 `j, then what must be the value of a?
2. If u = 3 `i + 5 `j - 2 `k, and we know that u dot v = 0 for the vector v = a `i + 10 `j + c `k, then what are the values of a and c?
3. Sketch the vectors <2, 5> + t * <3, -1> for t = 0, 1, 2 and 3, with the initial point of each vector at the origin. Sketch the line or smooth curve which passes through the terminal points of each of the four vectors. (i.e., each vector ends at a point; if you mark the four points, you want to sketch the line or curve through these points).
trig review; be able to sketch unit circle etc
break vectors into components etc
vector form a + t v (line, 2 dim then 3 dim)
condition a dot t v = 0 (perpendicular line, or perpendicular plane, depending on dimension)
... condition a dot (v x w) = 0 etc.
completing the square
all points above and below square, ellipse, circle, polygon ...
orthogonal means dot prod is zero (like perpendicular but in any dimension)
will need conic sections soon
The magnitude of a vector A = a_x i + a_y j in two-dimensional space is found using the Pythagorean Theorem to be sqrt(a_x^2 + a_y)^2.
[ The reasons for the above:In the x-y plane we find the magnitude of a_1 i + a_2 j to be sqrt(a_1 ^ 2 + a_2 ^ 2).The sum of this vector and the vector a_3 k is our original vector A = a_1 i + a_2 j + a_3 k.The vector a_3 k being perpendicular to the entire xy plane is also perpendicular to the vector a_1 i + a_2 j.So the vectors a_1 i + a_2 j and a_3 k are legs of a right triangle whose hypotenuse is therefore sqrt( (sqrt(a_1 ^ 2 + a_2 ^ 2) + a_3 ^2 ) = sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^ 2). ]
- If A = a_1 i + a_2 j + a_3 k andB = b_1 i + b_2 j + b_3 k, then A dot B = a_1 b_1 + a_2 b_2 + a_3 b_3.
-
A dot B = a_1 b_1 + a_2 b_2 + a_3 b_3
-
A dot B = || A || * || B || cos(theta).
cos(theta) = A dot B / || A || || B || so thattheta = arcCos( A dot B / || A || || B || ).
The cross product of two vectors A and B is a vector perpendicular to the two vectors, the direction determined by the right-hand rule, and its magnitude is || A || * || B || * sin(theta), where theta is the angle between the two.
The formula for the cross product is a little more complicated than the simple formula for the dot product, but it's not bad:A `X B is the determinant of the matrix whose first row is [ i, j, k ], whose second row is [ A_1, A_2, A_3 ], the coefficients of the A vector, and whose third row is [ B_1, B_2, B_3 ], the coefficients of the B vector.
v(t) = v_x(t) i + v_y(t) j = dx/dt i + dy/dt j = 2 t i + 3 t^2 j.
For example, at t = 2 our velocity vector is
v(2) = 2 * 2 i + 3 * 2^2 j = 4 i + 12 j.
A unit vector is a vector whose magnitude is 1. If you divide any vector by its magnitude you get a unit vector. The unit vector has magnitude 1 and its direction is the same as that of the original vector.
For example the magnitude of the vector 3 i + 4 j is sqrt(3^2 + 4^2) = 5. If you divide this vector by its magnitude you get the unit vector
u = (3 i + 4 j) / 5 = (3/5) i + (4/5) j = .6 i + .8 j.
This vector has magnitude 1, and its direction is the same as that of the original vector 3 i + 4 j.
As another example, if you divide the velocity vector v(2) = 4 i + 12 j, found just a little earlier, by its magnitude you get a unit vector.
The magnitude of v(2) = 4 i + 12 j is || v(2) || = sqrt( 4^2 + 12^2 ) = sqrt( 160 ) = 4 sqrt(10).
We give the unit vector in the direction of v(2) the name u(2):
u(2) = v(2) / || v(2) || = (4 i + 12 j) / (4 sqrt(10) ) = sqrt(10) / 10 * i + 3 sqrt(10) / 10 * j.
u(2) is the unit vector in the direction of the velocity, when t = 2.
**************
some extensions for next class (optional at this time):
The rate at which the velocity changes in the x direction is the derivative d v_x(t) / dt of the x velocity function v_x(t) = 2 t. The derivative of this function is the constant function 2. The derivative d v_y(t) / dt of the y velocity function v_y(t) = 3 t^2 is 6 t. So the x velocity changes at constant rate 2, while the y velocity changes at the increasing rate 6 t.
The derivative of our velocity function v(t) is therefore
dv / dt = 2 i + 6 t j.
The x component of the velocity increases steadily, while the y component increases at an increasing rate.
The general point ( x(t), y(t) ) can be represented by a vector which originates at the origin (0, 0) and extends to the point (x(t), y(t)). If we call this vector r(t), it is easy to see that
r(t) = x(t) i + y(t) j.
For our current example this vector is just
r(t) = t^2 i + t^3 j.
It makes perfect sense, and will later be shown to be the case, that the derivative of this r(t) vector is found by doing the obvious::
d r(t) / dt = 2 t i + 3 t^2 j.
We have simply treated the i and j vectors as constants.
In symbols, we say that the derivative of the vector function r(t) = x(t) i + y(t) j is found by taking the derivatives of the x(t) and y(t) functions, treating the i and j vectors as constants:
d r(t) / dt = d x(t) / dt * i + d y(t) / dt * j.
As we have seen, if x(t) = t^2 and y(t) = t^3, then d x(t) / dt = 2 t and d y(t) / dt = 3 t^2, so the derivative of our r(t) vector function is
d r(t) / dt = 2 t i + 3 t^2 j.
This is the velocity vector we found earlier:
v(t) = 2 t i + 3 t^2 j,
and illustrates the general relationship
v(t) = d r(t) / dt.
The tip of the r(t) vector traces out the curve we approximated previously.
The velocity vector v(t) is at every point tangent to the r(t) curve.
The rate of change of the velocity vector is the derivative we found earlier:
dv / dt = 2 i + 6 t j.
If we divide a vector by its magnitude, we get a vector of magnitude 1. We call a vector of magnitude 1 a unit vector.
In the present example we can divide the velocity vector by its magnitude. We obtain
v(t) / || v(t) || = ( 2 t i + 3 t^2 j ) / (sqrt( (2 t)^2 + (3 t^2)^2) ) = ( 2 t i + 3 t^2 j ) / (sqrt( 4 t^2 + 9 t^4 )).
This is a unit vector in the direction of the velocity. If we evaluate ( 2 t i + 3 t^2 j ) / (sqrt( 4 t^2 + 9 t^4 )) for any value of t, except t = 0 (where the denominator is zero), we will obtain a vector of magnitude 1.
***************** draft of 0823 below
The cross product of two vectors `A and `B is a vector perpendicular to the two vectors, the direction determined by the right-hand rule, and its magnitude is || `A || * || `B || * sin(theta), where theta is the angle between the two.
The formula for the cross product is a little more complicated than the simple formula for the dot product, but it's not bad:
`A `X `B is the determinant of the matrix whose first row is [ `i, `j, `k ], whose second row is [ `a_1, `a_2, `a_3 ], the coefficients of the `A vector, and whose third row is [ `b1, `b2, `b3 ], the coefficients of the `B vector.
`v(t) = v_x(t) `i + v_y(t) `j = dx/dt `i + dy/dt `j = 2 t `i + 3 t^2 `j.
For example, at t = 2 our velocity vector is
`v(2) = 2 * 2 `i + 3 * 2^2 `j = 4 `i + 12 `j.
A unit vector is a vector whose magnitude is 1. If you divide any vector by its magnitude you get a unit vector. The unit vector has magnitude 1 and its direction is the same as that of the original vector.
For example the magnitude of the vector 3 `i + 4 `j is sqrt(3^2 + 4^2) = 5. If you divide this vector by its magnitude you get the unit vector
`u = (3 `i + 4 `j) / 5 = (3/5) `i + (4/5) `j = .6 `i + .8 `j.
This vector has magnitude 1, and its direction is the same as that of the original vector 3 `i + 4 `j.
As another example, if you divide the velocity vector `v(2) = 4 `i + 12 `j, found just a little earlier, by its magnitude you get a unit vector.
The magnitude of `v(2) = 4 `i + 12 `j is || `v(2) || = sqrt( 4^2 + 12^2 ) = sqrt( 160 ) = 4 sqrt(10).
We give the unit vector in the direction of `v(2) the name `u(2):
`u(2) = `v(2) / || `v(2) || = (4 `i + 12 `j) / (4 sqrt(10) ) = sqrt(10) / 10 * `i + 3 sqrt(10) / 10 * `j.
`u(2) is the unit vector in the direction of the velocity, when t = 2.
The rate at which the velocity changes in the x direction is the derivative d v_x(t) / dt of the x velocity function v_x(t) = 2 t. The derivative of this function is the constant function 2. The derivative d v_y(t) / dt of the y velocity function v_y(t) = 3 t^2 is 6 t. So the x velocity changes at constant rate 2, while the y velocity changes at the increasing rate 6 t.
The derivative of our velocity function `v(t) is therefore
d`v / dt = 2 `i + 6 t `j.
The x component of the velocity increases steadily, while the y component increases at an increasing rate.
The general point ( x(t), y(t) ) can be represented by a vector which originates at the origin (0, 0) and extends to the point (x(t), y(t)). If we call this vector `r(t), it is easy to see that
`r(t) = x(t) `i + y(t) `j.
For our current example this vector is just
`r(t) = t^2 `i + t^3 `j.
It makes perfect sense, and will later be shown to be the case, that the derivative of this `r(t) vector is found by doing the obvious::
d `r(t) / dt = 2 t `i + 3 t^2 `j.
We have simply treated the `i and `j vectors as constants.
In symbols, we say that the derivative of the vector function `r(t) = x(t) `i + y(t) `j is found by taking the derivatives of the x(t) and y(t) functions, treating the `i and `j vectors as constants:
d `r(t) / dt = d x(t) / dt * `i + d y(t) / dt * `j.
As we have seen, if x(t) = t^2 and y(t) = t^3, then d x(t) / dt = 2 t and d y(t) / dt = 3 t^2, so the derivative of our `r(t) vector function is
d `r(t) / dt = 2 t `i + 3 t^2 `j.
This is the velocity vector we found earlier:
`v(t) = 2 t `i + 3 t^2 `j,
and illustrates the general relationship
`v(t) = d `r(t) / dt.
The tip of the r(t) vector traces out the curve we approximated previously.
The velocity vector `v(t) is at every point tangent to the `r(t) curve.
The rate of change of the velocity vector is the derivative we found earlier:
d`v / dt = 2 `i + 6 t `j.
If we divide a vector by its magnitude, we get a vector of magnitude 1. We call a vector of magnitude 1 a unit vector.
In the present example we can divide the velocity vector by its magnitude. We obtain
`v(t) / || `v(t) || = ( 2 t `i + 3 t^2 `j ) / (sqrt( (2 t)^2 + (3 t^2)^2) ) = ( 2 t `i + 3 t^2 `j ) / (sqrt( 4 t^2 + 9 t^4 )).
This is a unit vector in the direction of the velocity. If we evaluate ( 2 t `i + 3 t^2 `j ) / (sqrt( 4 t^2 + 9 t^4 )) for any value of t, except t = 0 (where the denominator is zero), we will obtain a vector of magnitude 1.