Assignment 2

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course Mth 277

6/15 1 pm

qa 09_2

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 magnitude of a vector `A = a_1 `i + a_2 `j + a_3 `k, in 3-dimensional space, is found by two applications of the Pythagorean Theorem to be sqrt( a_1 ^ 2 + a_2 ^ 2 + a_3 ^2).

A unit vector is a vector whose magnitude is 1. If you divide any vector by its magnitude you get a unit vector, and its direction is the same as that of the original vector.

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Question: `q001. Write an equation expressing the following statement:

The distance between the point (x, y, z) and the point (3, -4, 2) is 5.

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Your solution:

sqrt ( (3-x)^2 + (-4-y)^2 + (2-z)^2 ) = 5

(3-x)^2 + (-4-y)^2 + (2-z)^2 = 25

confidence rating #$&*:

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Given Solution:

The distance between the point (x, y, z) and the point (3, -4, 2) is sqrt((3 - x)^2 + (-4 - y)^2 + (2 - z)^2) = sqrt( (x - 3)^2 + (y + 4)^2 + (z - 2)^2).

This distance is 5 if and only if

sqrt( (x - 3)^2 + (y + 4)^2 + (z - 2)^2). = 5.

Squaring both sides we obtain

(x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25.

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Self-critique (if necessary):

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Question: `q002. Using the equation from the preceding, find the value of y if we know that x = 2 and z = 1.

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Your solution:

From the problem above:

(3-x)^2 + (-4-y)^2 + (2-z)^2 = 25

(3-2)^2 + (-4-y)^2 + (2-1)^2 = 25

(-4-y)^2 + 2 = 25

(-4-y)^2 = 23

-4-y = sqrt(23)

4+y = -sqrt(23)

y = -sqrt(23)-4

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If x = 2 and z = 1 then our equation

(x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25

becomes

(2 - 3)^2 + (y + 4)^2 + (1 - 2)^2) = 25,

or

1 + (y+4)^2 + 1 = 25

so that

(y + 4)^2 = 23

and

(y + 4) = +- sqrt(23).

Solving for y we get

y = sqrt(23) - 4

or

y = -sqrt(23) - 4.

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Given Solution:

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Self-critique (if necessary): Include +/- next time when dealing with square roots.

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Self-critique rating:

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Question: `q003. Using the equation from the first question, substitute z = 0. The resulting equation describes a circle. What are its center and radius?

Answer the same questions if you substitute z = 1 rather than z = 0.

Answer the same questions if you substitute z = -1 rather than z = 0.

Answer the same questions if you substitute z = -4 rather than z = 0.

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Your solution:

(3-x)^2 + (-4-y)^2 + (2-z)^2 = 25

If z = 0:

(3-x)^2 + (-4-y)^2 + 4 = 25

(x-3)^2 + (y+4)^2 = 21

The radius of the circle is sqrt(21) units, and the center is located at (3,-4).

If z = 1:

(x-3)^2 + (y+4)^2 + 1 = 25

(x-3)^2 + (y+4)^2 = 24

The radius of the circle is sqrt(24) units, and the center is located at (3, -4).

If z = -1:

(x-3)^2 + (y+4)^2 + 9 = 25

(x-3)^2 + (y+4)^2 = 16

The radius of the circle is 4 units, and the center is located at (3,-4).

If z = -4:

(x-3)^2 + (y+4)^2 + 36 = 25

(x-3)^2 + (y+4)^2 = -11

This circle cannot exist, since it has a negative radius.

confidence rating #$&*:

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Given Solution:

Our equation is

(x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25.

If z = 0 then our equation becomes

(x - 3)^2 + (y + 4)^2 + (0 - 2)^2) = 25,

which we easily rearrange to get

(x - 3)^2 + (y + 4)^2 = 21.

This is the equation of a circle in the x-y plane centered at (3, -4) and having radius sqrt(21), which is roughly 4.6.

If z = 1 then (z - 2)^2 becomes 1 and our equation becomes

(x - 3)^2 + (y + 4)^2 = 24.

This is the equation of a circle in the x-y plane centered at (3, -4) and having radius sqrt(24), which is roughly 4.9.

If z = -1 we get a circle of radius 6, centered at (3, -4).

If z = -4 we get the equation (x-3)^2 + (y+4)^2 = -11. Since squares can't be negative, there is no real-number solution for (x, y), and there will be no z = -4 point on this graph.

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Self-critique (if necessary):

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Question: `q004. Expand the equation you obtained in the first question by multiplying out the squares. Simplify into standard form, with all numbers and variable on the left and 0 on the right-hand side of the equation.

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Your solution:

(3-x)^2 + (-4-y)^2 + (2-z)^2 = 25

becomes:

9 - 6x + x^2 + 16 - 8y + y^2 + 4 - 2z + z^2 = 25

x^2 - 6x + y^2 - 8y + z^2 - 2z + 29 = 25

x^2 - 6x + y^2 - 8y + z^2 - 2z - 4 = 0

Confidence rating: 3

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Given Solution:

The equation

(x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25

becomes

x^2 - 6 x + 9 + y^2 + 8 y + 16 + z^2 - 4 z + 4 = 25

which in standard form is

x^2 - 6 x + y^2 + 8 y + z^2 - 4 z - 13 = 0.

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Self-critique (if necessary):

I'm not sure what I did wrong...

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Note that (2 - z)^2 = z^2 - 4 z + 4, not z^2 - 2 z + 4.

There was an error in the given solution.

x^2 - 6 x + 9 + y^2 + 8 y + 16 + z^2 - 4 z + 4 = 25

simplifies to

x^2 - 6 x + y^2 + 8 y + z^2 - 4 z + 4 = 0.

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Self-critique rating:

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Question: `q005. Is the vector 8 `i - 4 `j + 5 `k a multiple of the vector 4 `i + 2 `j - 5/2 `k?

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Your solution:

No, there is no scalar that can be multiplied to either to get the other.

confidence rating #$&*:

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Given Solution: The vector 8 `i - 4 `j + 5 `k is a multiple of the vector 4 `i + 2 `j - 5/2 `k if, and only if, there exists a constant c such that

8 `i - 4 `j + 5 `k = c ( 4 `i + 2 `j - 5/2 `k ).

Setting the `i, `j and `k components of the left-hand side respectively equal to the same components of the right-hand side we get the three equations

8 = 4 c

-4 = 2 c

5 = -5/2 c

The first yields solution c = 2. The last two yield solution c = -2.

c can't take both values at the same time, so the equations are not simultaneously true.

It follows that neither vector is a multiple of the other.

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Self-critique (if necessary):

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Question: `q006. How does your answer to the preceding determine whether or not the two vectors are parallel?

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Your solution:

If a vector is a multiple of another, the two vectors are parallel.

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Right idea but the statement you base the conclusion on needs to be stronger.

Had one vector been a multiple of the other, then your statement would allow us to conclude that the vectors are parallel.

However your statement doesn't address what happens if neither vector is a multiple of the other. That would require a statement equivalent to the following:

If two vectors are parallel, then each is a scalar multiple of the other.

We can combine the two statement into the single statement

Two vectors are scalar multiples of one another if and only if they are parallel.

This statement is equivalent to the statement in the given solution.

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Confidence rating:

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Given Solution: Two vectors are parallel if, and only if, one is a multiple of the other.

So the solution to the preceding shows that neither vector is a multiple of the other.

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Self-critique (if necessary):

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Question: `q007. What are the lengths of the sides of the triangle whose vertices are (4, 3 -2), (5, -1, 3) and (6, 4, 1)? Sketch a triangle, with its sides in the same proportion (as best you can in a few minutes without meticulously measuring everything). Based on your sketch does it seem plausible that this is a right triangle?

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Your solution:

Based on my sketch, it seems possible that the triangle with vertices given could be a right triangle, based on the angle formed at vertex (4,3,-2). However, the third dimension doesn't matter all that much since we're dealing with a 2D triangle.

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The first question posed here asked for the lengths.

Having found those lengths you would then sketch a triangle with its sides in the proportion of those lengths.

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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:"

&#Good responses. See my notes and let me know if you have questions. &#