Assignment 6

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course Mth 277

June 25 around 4:00 am

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.

qa 09_06

Section 9.6

Let P_0 = (2, 4, 5), P_1 = (4, 7, -2) and P_2 = (8, 1, -1). Let `u be the vector from P_0 to P_1 and `w the vector from P_0 to P_2.

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Question: `q001. Sketch the three points and the two vectors.

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Your solution:

P_0 = (2, 4, 5) P_1 = (4, 7, -2) P_2 = (8, 1, -1)

u = (4-2, 7-4, -2-5) = <2, 3, -7>

w = (8-2, 1-4, -1-5) = <6, -3, -6)>

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Question: `q002. Find the angle between `u and `w. Does the angle make sense in terms of your sketch?

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Your solution:

Theta = arcCos u dot v / || `u || || `v ||

u dot v = (2*6 + 3*(-3) + (-7)(-6) ) = 45

|| `u || = sqrt( 2^2 + 3^2 + (-7)^2 ) = sqrt(62)

|| `v || = sqrt ( 6^2 + (-3)^2 + (-6)^2 ) = sqrt(81) = 9

= arcCos (45 / sqrt(62) * 9) = arcCos(.64) = 50 degress

confidence rating #$&*:

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Given Solution:

P_0 = (2, 4, 5), P_1 = (4, 7, -2) and P_2 = (8, 1, -1).

`u = <4-2, 7-4, -2-5> = <2, 3, -7>

`v = < 6, -3, -6>

`u dot `v = 2 * 6 + 3 * (-3) + (-7) * ( - 6) = 12 - 9 + 42 = 45.

`u dot `v = || `u || || `v || cos(theta)

|| `u || = sqrt(2^2 + 3^2 + (-7)^2 ) = sqrt( 4 + 9 + 49) = sqrt(62)

|| ' v || = sqrt(6^2 + (-3)^2 + (-6)^2 ) = sqrt( 36 + 45 + 36) = sqrt(117).

Thus cos(theta) = `u dot `v / (|| `u || || `v ||) = 45 / sqrt( 62 * 117) = .54, very approximately, so

theta = arcCos(.54) = 56 degrees, again very approximately.

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Question: `q003. Find a vector perpendicular to both `u and `w. Call this vector the normal vector and denote it `n.

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Your solution:

u X w = <-39, -30, -24> or -39i - 30j -24k

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Given Solution:

The vector

`n = `u X `w = det ( `i, `j, `k; 2, 3, -7; 6, -3, -6 ) = -39 `i - 30 `j -24 `k, or <-39, -30, -24>

is perpendicular to both `u and `w.

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Question: `q004. Show that this vector `n is also perpendicular to the vector from P_1 to P_2.

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Your solution:

n = < -39, -30, -24>

v = P_1 to P_2 = < 4, -6, 1>

n dot v = -39(4) + (-30)(-6) + (-24)(1) = 0

confidence rating #$&*:

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Given Solution:

`n = -46 `i - 30 `j -26 `k, from the preceding.

P_1 = (4, 7, -2) and P_2 = (8, 1, -1).

The vector from P_1 to P_2 is < 4, -6, 1 >.

< 4, -6, 1 > dot <-39, -30, -24> = -39 * 4 - 30 * (-6) + 1 * (-24) = -156 + 180 + 24 = 0

So all three of the vectors between P0, P1 and P2 are perpendicular to the same vector.

This is as it should be, since the three points all lie in a common plane.

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Self-critique (if necessary):

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`q005. Let P = (x, y, z) be the coordinates of a point in space. What are the components of the vector from P_0 to P?

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Your solution:

P = (x, y, z)

P_0 = (2, 4, 5)

P_0 to P = < x-2, y-4, z-5 >

confidence rating #$&*:

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Given Solution:

From P_0 = (2, 4, 5) to P = (x, y, z) the components the vector is

< x - 2, y - 4, z - 5 >.

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Question: `q006. Write the condition that the vector from P_0 to P be perpendicular to `n. Express this condition as an equation.

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Your solution:

P_0 to P = < x-2, y-4, 5-z >

n = < -39, -30, -24 >

the dot product product between these vectors is = -39(x - 2) - 30(y - 4) - 24(z - 5)

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Given Solution:

From P_0 = (2, 4, 5) to P = (x, y, z) the components the vector are

< x - 2, y - 4, z - 5 >.

`n = <-39, -30, -24>

The dot product of these vectors is

-39 ( x - 2) - 30 ( y - 4) - 24 ( z - 5)

so the two vectors are perpendicular if and only if

-39 ( x - 2) - 30 ( y - 4) - 24 ( z - 5) = 0

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Question: `q007. Show that the coordinates of P_0 satisfy the equation found in the preceding, and also that the coordinates of P_1, as well as the coordinates of P_2, also satisfy this equation.

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Your solution:

= -39(x - 2) - 30(y - 4) -24(z - 5)

P_0 = (2, 4, 5)

= -39(2 - 2) - 30(4 - 4) -24(5-5) = 0

P_1 = (4, 7, -2)

= -39(4 - 2) - 30(7 - 4) - 24(-2 - 5) = 0

P_2 = (8, 1, -1)

= -39(8 - 2) - 30(1 - 4) - 24( -1 - 5) = 0

confidence rating #$&*:

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Given Solution:

The equation found in the following expressed the fact that the vector from P_0 to P = (x, y, z) is perpendicular to the normal vector:

-39 ( x - 2) - 30 ( y - 4) - 24 ( z - 5) = 0

The coordinates of P_0 are (2, 4, 5), so clearly if these are substituted for (x, y, z), the above equation holds.

P_1 = (4, 7, -2). Substituting these coordinates into the equation we get

-39 ( 4 - 2) - 30 ( 7 - 4 ) - 24 ( -2 - 5 ) = -78 - 90 + 168 = 0.

Your substitution of the coordinates of P_2 into the equation should similarly be shown to yield zero.

So the equation is satisfied by all three of our points.

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Question: `q008. If you have had linear algebra, show the condition that (x, y, z) satisfies your equation is equivalent to the condition that the vector from P_0 to P = (x, y, z) is in the span of the vectors `u and `v. This is also equivalent to the condition that `u and `v form a basis for the space of all such vectors.

If you haven't yet had linear algebra, you will later understand the question within this context, and should have no trouble understanding most steps of the given solution. The terminology of linear algebra, however, won't be required in this course.

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Your solution:

I never took linear algebra so I don't know how to show the condition.

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Given Solution:

Again the equation is

-39 ( x - 2) - 30 ( y - 4) - 24 ( z - 5) = 0.

Our vectors `u and `v are

`u = <2, 3, -7>

and

`v = < 6, -3, -6>.

The span of `u and `v is the set of all linear combinations of `u and `v, which can be understood whether or not you have had linear algebra to be the set of all vectors that can be obtained by adding multiples of `u to multiples of `v.

That is, the span of `u and `v is the set of all vectors of the form

c_1 * `u + c_2 * `v.

The vector from P_0 to P was found to be

The condition that this vector be in the span of `u and `v is therefore that, for some values of c_1 and c_2, the equation

c_1 `u + c_2 `v =

or

c_1 < 2, 3, -7> + c_2 < 6, -3, -6 > = .

If we eliminate c_1 and c_2 to solve this equation in terms of x, y and z we should find that we get our original equation.

You should proceed to solve this equation, if you haven't already done so, without reference to the solution given below (which presently does not appear to work out). You will set each component of the left-hand side equal to the corresponding component of the right-hand side, obtaining three equations in c_1, c_2, x, y and z.

As mentioned, the solution given below follows the right steps, but there is an apparent clerical error in some of the arithmetic:

Setting the components of the left- and right-hand sides equal we get the three equations

2 c_1 + 6 c_2 = x - 2

3 c_1 - 3 c_2 = y - 4

-7 c_1 - 6 c_2 = z - 5.

We can eliminate c_2 by adding 2 * second equation to first equation, obtaining

8 c_1 = x - 2 y - 10

We can also eliminate c_1 by adding the second equation to the first, obtaining

-5 c_1 = x + z - 7.

Solving both these equations for c_1, then setting the results equal, we get

c_1 = (x - 2 y - 10) / 8 = (-x - z + 7) / 5,

with the result that

5 ( x - 2 y - 10) = 8 ( -x - z + 7).

This should rearrange into the equation

-39 ( x - 2) - 30 ( y - 4) - 24 ( z - 5) = 0

but clearly does not.

What do you get when you complete the solution?

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Self-critique (if necessary):

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Question: `q009. Write down the condition that the vector from (x0, y0, z0) to (x, y, z) is perpendicular to the vector `n = a `i + b `j + c `k, and simplify your condition into an equation.

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Your solution:

( x0, y0, z0) to (x, y, z)

= < x - x0, y - y0, z - z0 >

< x - x0, y - y0, z - z0 > dot < a, b, c >

= a(x - x0) + b(y - y0) + c(z - z0) = 0

confidence rating #$&*:

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Given Solution:

The vector from (x0, y0, z0) to (x, y, z) is , and is perpendicular to the vector `n = a `i + b `j + c `k provided

dot < a, b, c > = 0

so that

a ( x - x0) + b ( y - y0) + c ( z - z0) = 0.

This is the desired equation, and is the equation of the plane through (x0, y0, z0) perpendicular to the vector a `i + b `j + c k.

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Self-critique (if necessary):

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Question: `q010. Find the coordinates of a point (x, y, z) which satisfy the condition that 2 x + 3 y - 5 z = 30.

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Your solution:

point ( x, y, z) 2 x + 3 y - 5 z = 30

let x & y = 0

-5z = 30

z = -6 ( 0, 0, -6)

let y & z = 0

2x = 30

x = 15 (15, 0, 0)

let x & z = 0

3y = 30

y = 10 (0, 10, 0)

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Given Solution:

The equation has three variables x, y and z. We should be able to pick the values of two of these variables and solve for the third.

For example let x = 5 and y = 7. Our equation becomes

2 * 5 + 3 * 7 - 5 z = 30

We easily solve for z to get

z = 1/5.

The point (5, 7, 1/5) therefore satisfies the equation.

We could of course let two of our unknowns be 0.

Letting x and y both be zero we find that -5 z = 30 so z = -6 and our point is (0, 0, -6).

Letting x and z both be zero we find that 3 y = 30 so y = 10 and our point is (0, 10, 0)

Letting z and y both be zero we similarly obtain the point (15, 0, 0).

Plotting these three points on a set of x y z axes, we get a fairly good picture of the plane defined by the equation.

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Self-critique (if necessary):

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Self-critique rating:

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&#Very good responses. Let me know if you have questions. &#