091 Query

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course MTH 277

9/12 9:02

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Question: Sketch the vector from P to Q, write it into standard component form, and find ||PQ||. P=(4,-1) Q=(-3,7).

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Your solution:

PQ = (4 + (-3), -1 + 7)

PQ = (1,6)

||PQ|| = sqrt ((1)^2 + 6^2)

||PQ|| = sqrt (37)

||PQ|| ≈ 6.0828

@& Whatever program you used to write this used a symbol that's not part of the ASCII set.

I do have this program set to interpret things like ≈ into text, but I'm not sure what that symbol is. Most likely an = sign. Let me know if I'm right and if so it will take me just seconds to permanently add this to the routine.*@

confidence rating #$&*:232;2

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Given Solution:

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Self-critique (if necessary):

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Question: Let u = <-4,3> and v = <2,-1/2>. Find scalars s and t so that s * <0,3> + tu = v.

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Your solution:

s * <0.3> + t * <-4,3> = <2, -1/2>

t = -1/2 s * <0,3> + -1/2 * <-4,3> = <2, -1/2>

s * <0,3> + <2, -3/2> = <2, -1/2>

s = 1/3 1/3 * <0,3> + <2, -3/2> = <2, -1/2>

<0,1> + <2, -3/2> = <2, -1/2>

<0 + 2,1 + -3/2> = <2, -1/2)

<2, -1/2> = <2, -1/2>

@& Note how this question translates into the two equations

-4 t = 2

3 s + 3 t = -1/2

The solution is as you say.*@

confidence rating #$&*:

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Given Solution:

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Question: Let u = 4i - 3j, v = -3i + 4j , and w = 6i - 3j. Write the expression ||u|| ||v|| w in standard form.

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Your solution:

||u|| = sqrt (4^2 + (-3)^2)

||u|| = sqrt (16 + 9)

||u|| = sqrt (25)

||u|| = 5

||v|| = sqrt ((-3)^2 + 4^2)

||v|| = sqrt (9 + 16)

||v|| = sqrt (25)

||v|| = 5

||u|| ||v|| w= 5 * 5 * (6i - 3j)

||u|| ||v|| w= 25 * (6i - 3j)

||u|| ||v|| w= 150i - 75j

confidence rating #$&*:

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Given Solution:

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Question: Let u = 4i + j, v = 4i + 3j, w = -i + 2j. Find a vector of length 3 with the same direction as u - 2v + 2w.

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Your solution:

(4i + j) - 2(4i + 3j) + 2(-i + 2j)

4i + j - 8i - 6j - 2i + 4j

-6i - j

s= 1/3 * (-6i - j) s= vector of length 3

s= -2i -1/3j

@& The length of -6 `i - `j is sqrt(37), so a unit vector in the direciton of this vector is

-6 sqrt(37) / 37 `i - sqrt(37) / 37 `j.

A vector of length 3 in this direction is

3 * (-6 sqrt(37) / 37 `i - sqrt(37) / 37 `j)

The vector -2 `i - 1/3 `j is not of length 3.*@

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Given Solution:

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Question: Show that the vector v = cos(theta)i + sin(theta)j is a unit vector for any angle theta.

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Your solution:

||v|| = sqrt ( (cos theta)^2 + (sin theta)^2)

For any angle theta, (cos theta)^2 + (sin theta)^2 = 1

||v|| = sqrt (1)

||v|| = 1

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

@& Very good. You did have one error, and I inserted what should be a useful note on one other problem.

Let me know if you have questions.*@