Calculus III - Unit 12
how do you graph a plane in R^3?
- All you are doing for this is finding the points of intersection between the plane and the three axes. 1. Set y and z = 0. Solve for x. Plot this point as x = #. 2. Follow this same step to solve for the y and z points to plot. 3. Once all three points are plotted, connect them with a triangle. 4. You may extend the lines further.
PROPERTIES OF CROSS PRODUCTS A x B =
- B x A
parallelogram method
- Completely form a parallelogram using the vectors - The diagonal of this figure is the resultant vector
Plotting a circle in R^3
- Given that your equation only involves x and y coordinates (x^2 + y^2 = 1) we know that z can be any value. - Plot the circle normally in the x-y plane with a radius of one. - Draw lines going upward from the circle (parallel to the z axis) to signify that z can be any value. Remember that when a letter (either x y or z) is not included in an equation, it's value is zero on the graph.
Plotting singular letter points in R^3
- If you have an x = #, y and z can be any number. If you have y = #, x and z can be any number. If you have z = #, x and y can be any number. - These types of problems will always result in a plane that has sides parallel to the two letters (x,y, or z) that can be filled by any number.
unit vector definition
- Unit vectors are vectors with a length of one. We will often scale a vector to a unit vector - Unit vectors can be found by computing the magnitude/length of a vector
DOT PRODUCT PROPERTIES 0 * A =
0
PROPERTIES OF VECTORS a + -a =
0
PROPERTIES OF VECTORS c (a + b) =
ca + cb
PROPERTIES OF VECTORS (c + d) a =
ca + cd
find two different unit vectors orthogonal to A and B
can do A x B/ I A x BI and B x A/ I B x AI this really is just saying to find one unit vector (by using the cross product) and then negating it to determine the second unit vector.
the negation of a vector orthogonal to two vectors
is still orthogonal to the two vectors (just pointing in the opposite direction)
column vector
listed in a vertical form
row vector
listed in horizontal form
scalar
magnitude
to find the projection....
multiply the unit vector (magnitude/length) by the component (scalar)
another name for length of vector
norm of vector
the normal vector of a plane is __________ to that plane
orthogonal (perpendicular)
If a line's direction vector has a cross product of zero with a plane's normal vector, that line is _________ to the plane
parallel
if two lines do not intersect they must be
parallel or skew
Since the lines are skew, they lie on
parallel planes with the normal vector V1 x V2 = N.
If a line has the same direction vector as the normal vector of the plane, that line is _________ to the plane
perpendicular
normal is another word for
perpendicular
comp a B (in reference to proj a B) =
plus/minus IProj a BI
vectors are used to represent
points
parametric equation of a line
r(t) = r0 + tv
decomposition of vectors into standard basis vectors
remember, the standard basis vectors are E1, E2, and E3 (or i j k). All you have to do to decompose a vector into the standard basis vectors is to write the vector as coefficients of E1,E2,E3 and then subsequently i, j, k.
A x B follows the
right hand rule point your fingers in the direction of A curl your fingers in the direction of B where ever your thumb ends up pointing is the direction of the cross product
Coordinates in R^3 are
right handed
the component of B in the direction of A is a
scalar
you can multiply vectors by a
scalar (and get longer or shorter versions of that vector)
first octant
set of points whose coordinates are all positive
how do we alter the problem to solve for the distance from a line to a skew line?
solve for the distance between a line and a plane after altering the problem
<i, j, k> are the
standard basis vectors
what are the two vector addition methods?
tail to tip method and parallelogram method
I A x BI =
the area of a parallelogram generated by A and B
scalar projection is also known as
the component of ______ along ________.
coordinate planes
the three coordinate axes create coordinate planes (xy, yz, and xz planes).
octant
the three coordinate planes divide space into eight parts. The first octant in the foreground is determined by positive axes.
parallel
the vector directions are scalar multiples of each other (have to be)
Vector Projection
the vector produced when one vector is resolved into two component vectors, one that is parallel to the 2nd vector and one that is perpendicular to the 2nd vector
if three points are not on a line
they are in triangle
if you are asked to find something orthogonal to two other vectors
this is a hint to use the cross product. the result of the cross product is always perpendicular to the two original vectors
right hand rule
this rule determines the orientation of the z axis. If you curl the fingers of your right hand around the z-axis in the direction of a 90 degree counterclockwise rotation from the positive x-axis to the positive y-axis, then your thumb points in the positive direction of the z-axis
coordinate axis
three lines (represented by x, y, and z) that are perpendicular to each other
unit vector formula
v/||v|| this represents magnitude/length
A point, Q, is on the plane if
vector PQ is perpendicular to N or if the dot product of PQ and N = 0.
to add vectors gemoetrically
we can slide them around in space
z = 0 is the
xy plane
y = 0 is the
xz plane
can a component be negative?
yes
X = 0 is the
yz plane
distance from a point to a plane equation
D = I N * QPI / INI
PROPERTIES OF CROSS PRODUCTS (cA) x B =
c (A x B) = A x (cB)
PROPERTIES OF VECTORS (cd) a =
c (da)
Proj a B =
(A * B/ IAI^2) times A
PROPERTIES OF CROSS PRODUCTS A x (B x C) =
(A *C)B - (A* B)C
PROPERTIES OF CROSS PRODUCTS A * (B x C) =
(A x B) * C
PROPERTIES OF VECTORS a + (b + c) =
(a + b) + c
Equation of a sphere centered at (h, k, l) with radius r
(x-h)^2+(y-k)^2+(z-l)^2=r^2 can also be written as sqrt [(x-h)^2+(y-k)^2+(z-l)^2] = r
symmetric equation of a line
(x-x0)/a = (y-y0)/a = (z-z0)/a
PROPERTIES OF CROSS PRODUCTS A * ( A x B) =
0 if you think about this logically, the result of a cross product is perpendicular to both the starting vectors. And, since the dot product of two perpendicular vectors is zero, this must always be zero.
Determining if points are collinear
1. Find the distances between the points 2. If two of the distances add up to the third distance, the points are collinear. - If the points are not collinear, they are probably in a triangle formation
finding the equation of intersection for two planes.
1. First, make sure that the planes are NOT parallel (look at their normal vectors). In order to do so, make sure they are not scalar products of each other OR make sure their cross product does not equal zero. 2. Take the cross product of their normal vectors. This results in a line parallel to the planes. This is the direction vector for the line of intersection. 3. Create a system of equations using the two plane equations. Set z = 0. (if z = 0 does not work, try x and y =0). 4. Solve the system of equations for each variable. 5. Use this point in your equation of the line.
Finding the Equation of a plane through 3 given points:
1. Form two vectors from the given points (subtract points) 2. Find the cross product of these vectors in order to find the normal vector. 3. Use this normal vector in the plane equation. 4. Substitute one of the original points into your equation to solve for D.
Finding the Equation of a sphere using complete the square:
1. Group all of the like-variables together (xs with xs, ys with ys, zs with zs). Group by ( ). 2. Take (b/2)^2 and add it inside the ( ) while subtracting the same value from outside the ( ). 3. Simplify the inside of ( ) to have a squared sign. 4. Determine the center of the sphere (remember to flip the signs!) 5. Determine the radius by adding up all stray numbers and moving them to the right of the =. 6. The radius is the square root of step five.
Finding the Distance between two planes
1. Make sure that the planes are parallel. 2. Set y and z equal to zero for one of the planes and solve for x. 3. Create a point from this (x, 0, 0). This is a point on the first plane. 4. Calculate the distance from the point on plane one to plane 2 using the previous method. look on sg for formula
Decomposing a vector into its component vectors
1. Put the vector coordinates in front of E1, E2, E3 as if they were coefficients (remember, this is just like multiplying by 1) 2. This is equivalent to the coefficient in front of i, j, and k
finding the point of intersection between two lines
1. Set each component (x, y, and z) from the two equations equal to each other. 2. Replace the variables on one side of the equal sign with a different letter from the first side. 3. We cannot assume that these two equations have "t" representing the same number. Use s on one side. 4. Use the system of equations to eliminate a variable and solve for the other. 5. Solve for the remaining variable. 6. Plug the variable values back into the original equations presented to solve for the point of intersection. If there is a solution to this process, the line intersects. If there is not a solution to this process, the lines do not intersect and we must have either skew or parallel lines.
Plotting points (x, y, z) in R^3
1. Start with the x coordinate 2. Then, from that x coordinate, move over in the y-direction the designated amount of spaces. 3. Then, from the result of step two, move in the z-direction the designated amount of spaces.
finding the distance between a line and a skew line
1. u sing the two planes, write down their normal vectors. 2. Cross the normal vectors to find the normal vector of the two. - This normal vector is perpendicular to both lines because the skew lines must be in different planes. 3. Find a point on both p1 and p2 (use the points given to you in the line equations). 4. Subtract the points on p1 and p2 to find a vector between the two (P1P2) 5. Compute the scalar projection (component) of P1P2 onto the normal vector, n. 6. This gives you your distance between the skew lines as the component vector is a scalar value.
k = E3 =
<0, 0, 1>
j = E2 =
<0, 1, 0>
i = E1 =
<1, 0, 0>
DOT PRODUCT PROPERTIES (cA) * B
= c (A * B) = A * (cB)
equation of a plane
A (x-x0) + B(y - y0) + C(z -z0) = 0 Ax + By + Cz = D
comp a B =
A * B/ IAI
DOT PRODUCT PROPERTIES A * (B + C) =
A *B + A *C
finding the angle between two vectors using the dot product
A *B = IAI IBI cos theta
dot product
A B = a1b1 + a2b2 +a3b3
tail to tip method
To add vectors geometrically, remember that vectors are allowed to slide around in space (vectors remain parallel to the original vector and pointing in the same direction)... - Place the tail of one vector on the head (arrow) of the other
if the dot product of A * B = 0
A is perpendicular to B
PROPERTIES OF CROSS PRODUCTS A x (B + C) =
A x B + A x C
using the cross product to determine an angle between two vectors
A x B = IAI IBI sin theta
PROPERTIES OF CROSS PRODUCTS (A + B) x C =
A x C + B x C
Finding the Angle Between Vectors using the DOT PRODUCT
A* B = IAI IBI cos theta
DOT PRODUCT PROPERTIES A * B =
B *A
orthogonal projection of B onto A: Orthoga B =
B - Proj a B
Finding the angle of intersection between two intersecting lines formula
Cos theta = V1 V2/ IV1I IV2I
How is distance denoted using the letters of the points
Denoted with absolute value bars around the product of the two lettered points.
i is also denoted as
E1
j is also denoted as
E2
k is also denoted as
E3
finding the equation of a line from two points
Find a vector by subtracting the two points Use one of the points as the point in the equation
for a 3x3 matrix:
For a 3 x 3 matrix, expand along a row to lower dimensional determinants.
formula for finding the distance between a line and a skew line
I N * P1P2I / INI
Finding the distance from a point to a plane
I N * QPI / INI
Volume of a parallelepiped
IA*(BxC) I = Ia1 a2 a3I I b1 b2 b3I I c1 c2 c3I
length of vector equation
IAI = sqrt [x^2+y^2+z^2]
a different component formula (less used) Comp a B =
IAI cos theta
DOT PRODUCT PROPERTIES A * A =
IAI^2
the length (norm) of a vector is denoted by
IIAII or IAI
when writing the equation of a line....there are...
KEEP IN MIND THAT THERE ARE MANY DIFFERENT EQUATIONS THAT REPRESENT THE SAME EXACT LINE
skew lines
Lines that do not intersect and are not coplanar - in parallel planes
Distance Formula (for two points)
P1P2 = sqrt [(x1-x2)^2+(y1-y2)^2+(z1-z2)^2]
plane equations go through
Plane equations - through a point normal to the vector N <A, B, C>
subtracting vectors
Subtract the coordinating coordinate points P1P2 = <x2 - x1, y2 - y1, z2 - z1>
DO NOT FORGET WITH CROSS PRODUCTS...
THE J TERM IS ALWAYS NEGATED Do not forget that the j term in (i, j, k) is always negated (subtracted) when adding the determinants together
finding the distance from a point to a line
Use properties of cross products D = I QP x VI / IVI Here, P is the given point and Q is a point on the line.
vector addition
Vector addition -> literally just add the corresponding coordinates - Keep in mind: you cannot add vectors if they do not have the same amount of coordinates
PROPERTIES OF VECTORS 1a =
a
PROPERTIES OF VECTORS a + 0 =
a
if you flip the vector around you get
a negative vector
Z = k represents
a plane parallel to the xy plane
Y = k represents
a plane parallel to the xz plane
X = k represents
a plane parallel to the yz-plane
the absolute value of a vector gives you
a scalar
sphere definition
a set of points a fixed distance, r, from a center P (h, k, l)
vector projection of B onto A (in words)
a vector in the direction of A
displacement vector
a vector that measures the distance between two points
position vector
a vector that starts from the origin and measures to a point
PROPERTIES OF VECTORS a + b =
b + a
how do you solve a determinant?
cross multiply and subtract the diagonals
scalar multiplication
distribute the scalar to all coordinates of the vector
the two sides of a triangle added are always
greater than the third side
cross product expanded form
i (a2b3 - a3b2) + j(a3b1 - a1b3) + k(a1b2 - a2b1)
projection
if we drop a perpendicular from a point to a plane, we get this. - The coordinates of the end point of the projection should be the same as the original point BUT there is a zero in place of one number. That number is the value for the coordinate that is not one of the axes used in that projection (if the projection ends in the yz plane, the x value of the coordinate for this projection is 0).
surface
in R3, an equation with x,y,z represents a surface
Planes can either
intersect or be parallel. they cannot be skew.
