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course Mth 277
Sorry for the late work.12 am 9/26/12
120912
Today's class and the questions below are relevant to the following:
Assignment 06: q_a_09.6, Text Section 9.6, Problem Set 9.6, query_09.6
Assignment 07: q_a_09.7, Text Section 9.7, Problem Set 9.7, query_09.7
`q001. Show that any vector from one of the points (2, 1, 3), (3, 2, 5), (-2, 1, 4) to any other point is perpendicular to the cross product of any two such vectors (provided the cross product is not zero).
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(2, 1, 3) to (3, 2, 5). i + j + 2k.
(2, 1, 3) X (3, 2, 5). -i -j +k.
-i -j +k dot i + j + 2k. -1 -1 + 2 = 0. Thus, perpendicular.
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`q002. Write as an equation: The vector from (x, y, z) to (1, 4, -5) is perpendicular to the vector 2 `i + 4 `j - `k.
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(1-x)i + (4-y)j + (-5-z)k X 2i + 4j - k = 0.
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`q003. Give a convincing argument that the set of all points (x, y, z) which satisfy the conditions of the preceding form a plane in 3-dimensional space.
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Well, if these three points (whatever you plug in there) are always orthogonal to vector 2i + 4j - k , that is really the definition of a plane. You have a vector, with an outsweeping set of orthogonal points that constitute a plane.
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`q004. The equations 2 x + 4 y - z + 12 = 0 and 3 x + 2 y + 4 z = 0 represent planes in space. Find a point on each and a unit vector perpendicular to each. Find also a unit vector parallel to both planes.
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For plane 2 x + 4 y - z + 12 = 0.
point (0,0 12).
Normal vector = 2i + 4j -k. Unit vector of this = (2i + 4j -k)/sqrt(21).
Vector parallel to plane. 2i + 4j -k dot (xi + yj + zk) = 0.
For plane 3 x + 2 y + 4 z = 0.
point (0, 0, 0) or (0, 2, -1)
Normal vector= 3i + 2j + 4k. Unit vector of this= (3i + 2j + 4k) / sqrt(29).
Vector parallel to plane. 3i + 2j + 4k dot (xi + yj + zk) = 0.
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`q005. Find the intersection of the planes in the preceding in two ways:
Treat the two equations as simultaneous equations. Eliminate z to get an equation in the x-y plane. Parameterize this equation. Then find a corresponding parameterization of z.
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2 x + 4 y + 12 = 0
3 x + 2 y = 0
x= 3, y = -4.5.
2 x + 4 y - z + 12 = 0. z= 24.
3 x + 2 y + 4 z = 0.
Having a hard time with this.
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2 x + 4 y - z + 12 = 0.
3 x + 2 y + 4 z = 0.
First equation: z = 2 x + 4 y + 12
Second equation: 3 x + 2 y + 4 ( 2 x + 4 y + 12) = 0
11 x + 18 y + 48 = 0
y = -11/18 x -48/18 = -11/18 x - 24/9
x = t
y = -11/18 t - 24/9
z = 2x + 4 y + 12 = ... (substitute expressions x = t and y = -11/18 t - 24/9)
There's a parameterization.
Read off a direction vector and find a point (e.g., let t = 0 and you get the point (0, -24/9, ...); you can fill in the ... ).
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Find a vector `v which is parallel to both planes, and a single point at which the planes intersect (one suggestion: find the line of intersection of each plane with the xy plane and then find the intersection of these two lines). The intersection of the two planes is the line through this point, parallel to the vector `v.
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2 x + 4 y - z + 12 = 0
3 x + 2 y + 4 z = 0
Two normal vectors 2i + 4j -k and 3i + 2j + 4k. If you find a vector orthogonal to both normal vectors, that vector must be parallel to the planes.
(2i + 4j -k) X (3i + 2j + 4k).
18i - 11j - 8k.
Having a hard time with this.
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If correct, 18 `i - 11 `j - 8 `k is your direction vector. Find a point on both planes (e.g., let x = 0 and solve the resulting two simultaneous equations for y and z), then construct the line in the usual manner.
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`q006. For the plane through (1, 4, -5) and perpendicular to the vector 2 `i + 4 `j - `k, find the magnitude of the projection of the vector from (5, 8, 3) to (1, 4, -5) on the vector 2 `i + 4 `j - `k.
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Scalar projection of v onto w = ( v dot w )/ ||w||.
-4 i -4j - 8k dot 2 `i + 4 `j - `k. -8 -16 +8 = -16.
||w|| = sqrt(21).
-16/ sqrt(21).
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Find the coordinates of another point on the same plane. Find the magnitude of the projection of the vector from (5, 8, 3) to this new point on the vector 2 `i + 4 `j - `k.
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(1, 4, -5) and perpendicular to the vector 2 `i + 4 `j - `k.
Point (2, 5, 1) to (5, 8, 3). This is vector 3i + 3j + 2k.
v= 3i + 3j + 2k projected onto w= 2 `i + 4 `j - `k. Scalar projection of v onto w = ( v dot w )/ ||w||.
(6+12-2)/sqrt(21).
16/sqrt(21).
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Find the magnitude of the projection of the same vector on a unit vector parallel to 2 `i + 4 `j - `k.
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u = (2 `i + 4 `j - `k) / sqrt(21). .436i + .873j - .218k.
v= 3i + 3j + 2k projected onto w= .436i + .873j - .218k. Scalar projection of v onto w = ( v dot w )/ ||w||.
3.491/.999...
3.49.
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`q007. Find the trace of the quadric surface x^2 + 4 y^2 + z^2 = 100 in the plane z = c.
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x^2 + 4y^2 + c^2 = 100.
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That's an ellipse. Should be put into standard form:
x^2 + 4 y^2 = 100 - c^2 so
x^2 / (sqrt(100 - c^2) ) + y^2 / sqrt(25 - c^2 / 4) = 1.
Semiaxes are
(sqrt(100 - c^2) ) and sqrt(25 - c^2 / 4) .
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How does this trace change with the value of c?
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x^2 + 4y^2 = 100 - c^2.
This is an ellipse, as c gets larger, the a and b values of the ellipse get smaller, so the ellipse shrinks.
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Are there limits on the value of c for which the trace consists of at least one point?
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C cannot be equal to or greater than 10 otherwise its not an ellipse. Neither can it be equal to or smaller than -10. Domain of c = (-10, 10).
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Explain how the behavior of the traces allows you to understand the shape of the surface.
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A trace in different planes allows essentially a cross-sectioned view of whatever three-dimensional object you're examining. By examining this object in multiple planes, you can piece together a wire-frame image/outline of what the object should look like.
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Repeat for the surface x^2 + 4 y^2 - z^2 = 100, and contrast the results with those you obtained above.
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Hyperboloid of one sheet.
x^2 + 4 y^2 - c^2 = 100.
x^2 + 4 y^2 = 100 + c^2. As c increases, a and b increase in the ellipse shape in the xy plane. Thus, the ellipse gets bigger.
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`q008. For the surface x^2 + 4 y^2 - z^2 = 100, find the trace in the plane y = c.
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x^2 + 4 c^2 - z^2 = 100
x^2 - z^2 = 100 - 4 c^2
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How does this trace change with the value of c?
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Hyperbola in the x z planes. As c gets larger, a and b of the hyperbola get smaller, causing the hyperbola to shrink.
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Are there limits on the value of c for which the trace consists of at least one point?
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Domain of c ( -sqrt(25), sqrt(25) ).
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Explain how the behavior of the traces allows you to understand the shape of the surface.
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A trace in different planes allows essentially a cross-sectioned view of whatever three-dimensional object you're examining. By examining this object in multiple planes, you can piece together a wire-frame image/outline of what the object should look like.
By knowing that this is an ellipse in the xy plane and a hyperbola in the xz plane we can determine that this is a hyperboloid of one sheet.
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`q009. Find the intersection of the surface x^2 + 4 y^2 - z = 100 with the plane 2 x + 3 y + 4 z = 12. Suggestion: Eliminate z, parameterize the resulting equation in x and y, then in terms of this parameterization find a parameterization for z.
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Parameterized line must be parallel to the plane 2 x + 3 y + 4 z = 12, and therefore to the vector 2i + 3j + 4z. This is direction vector.
When z = 0, x^2 + 4 y^2 = 100 and 2 x + 3 y = 12.
2 x + 3 y = 12. Contains point (3, 2, 0)
t = (x - 3)/2 = (y-2)/3 = (z-0)/4.
x= 2t+3. y = 3t + 2.
Plug into x^2 + 4 y^2 = 100. You get t = .8112 and -2.3112.
Again, feel like I'm close, but I can't quite get there.
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2 x + 3 y + 4 z = 12
so
z = 3 - x/2 - 3/4 y.
Substituting into the equation for the surface we get
x^2 + 4 y^2 + x/2 + 3/4 y = 103.
Complete the squares on x and y to get the form
(x - h)^2 + 4 (y - k)^2 = c
This is an elllipse in the xy plane. Parameterize it. Note that an ellipse centered at (h, k) can be parameterized by
x = h + a cos(theta)
y = k + b cos(theta)
where a and b are the semiaxes.
Then substitute these parametric equations for x and y into z = 3 - x/2 - 3/4 y. Bingo, you've got a parameterization of the intersection.
You do need to figure out an appropriate domain for theta.
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Check my notes. Questions 1-4 were on a previous posting.
As you know I love to answer questions.
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