query 6

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course Mth 277

2/8 2

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

query_09_6

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Question: Write the equation of the plane 3(x-2) - 2(y-1) - 3(z-5) = 0 in standard form.

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Your solution:

Standard Form

3x - 2y - 3z +11 = 0

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Given Solution:

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Self-critique (if necessary):

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Question: Find the equation of the plane containing the point P(-1,3,2) and having normal vector N = 3j - 1k.

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Your solution:

Direction numbers

[0, 3, -1]

Parametric Form

x = -1, y = 3 + 3t, z = 2 - t

Equation of the plane

3(3 + 3t) - (2 - t) = 0

t = -7/10

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Given Solution:

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Self-critique (if necessary):

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Question: Find two unit vectors perpendicular to the plane x + 3y - 4z = 2.

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Your solution:

??????Could use some help with this one, I believe its centered around the normal vector????????

@& The normal vector is perpendicular to the plane.

If you divide it by its magnitude you get a unit vector perpendicular to the plane.

If you then take the negative of this vector it's still perpendicular, and still a normal vector.

That gives you two.*@

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Given Solution:

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Self-critique (if necessary):

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Question: Find the distance between the point (-1,2,1) and the plane which contains the point (3,3,-2) and is normal to the vector N = -2i + j + 3k.

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Your solution:

Distance = |QP dot N| / ||N||

QP dot N = 16

||N|| = `sqrt(14)

Distance = 16/`sqrt(14)

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Given Solution:

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Question: Find the distance between the lines (x+1)/-2 = (y+2)-2 = (z+1)/-1 and (x-4)/5 = (y+1)/2 = (z-1)/3

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Your solution:

L_1: x = -1 - 2s, y = -2 - 2s, z = -1 - s

L_2: x = 4 + 5t, y = -1 + 2t, z = 1 + 3t

Find pt Q on plane#1 by setting s = 0

x = -1, y = -2, z = -1, so Q(1, -2,-1) is our pt.

We have vectors

v_1 = <-2, -2, -1> and v_2 = <5, 2, 3>, which are parallel to L_1 and L_2

Taking cross product of v_1 and v_2

N = V_1 X v_2 = -4i + j + 6k

Letting t = 0 we see the pt (4, -1, 1), so equation for plane#2

-4(x - 4) + (y + 1) + 6(z - 1) = 0

= -4x + y + 6z - 11 = 0

Dist from L_1 to L_2 is the same as Q(1, -2, -1) to the plane#2 so,

Distance = |-4(1) + (-2) + 6(-1) - 11| / `sqrt(-4^2 + 1^2 + 6^2)

= 23/`sqrt(53)

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Given Solution:

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Question: Find the equation of the sphere with center C(-2,7,1) and tangent the the plane x + 4y - 2z = 10.

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Your solution:

First we find radius

r = |(-2 + 4(7) - 2(1) - 10) / `sqrt(1^2 + 4^2 + (-2)^2)| = |14/`sqrt(21)|

Equation of the sphere

(x + 2)^2 + (y - )^2 + (z - 1)^2 = (14/`sqrt(21))^2

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Given Solution:

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Self-critique (if necessary):

@& The distance from the plane to the center will be the radius of the sphere.

The point (0, 0, 5) is on the plane.

The vector from (-2, 7, 1) to (0, 0, 5) is <2, -7, -6>.

Dot this with a unit normal vector and we get the distance of (-2, 7, 1) from the sphere -- i.e., the distance from the center of the sphere to the tangent plane, and hence the radius of the sphere.

A normal vector is <1, 4, -2> so a unit normal is <1/sqrt(21), 4 / sqrt(21), -2 / sqrt(21)>.

<2, -7, -6> dot <1/sqrt(21), 4 / sqrt(21), -2 / sqrt(21)> = (2 - 28 + 12) / sqrt(21) = -14 / sqrt(21), so the equation of the sphere is

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