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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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X should be dot
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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.
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There are infinitely many solutions for (x, y, z)
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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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Infinitely many solutions here as well.
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We still don't have a vector parallel to both planes.
We have an equation for a vector parallel to the first plane, and an equation for a vector parallel to the second, but no vector that satisfies both.
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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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Almost midnight and I want to post everything up to here. Will continue this in the next posting.
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