#$&* course Mth 277 9/13/2011 @ 3:15 p.m. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Let u = <-4,3> and v = <2,-1/2>. Find scalars s and t so that s * <0,3> + tu = v. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1/3(0i + 3j) + (-1/2)(-4i + 3j) = (2i - 1/2j). Since the “first” i component is 0i, any s will still give us 0i. This means that -4i * t has to give us 2i, which means t = -1/2. If t = -1/2, then -1/2 * 3j gives us -1 and ½ j. To get the -1/2 we need for the solution, we must add 1. To get the 1, we multiply 3j by 1/3, which makes our s = 1/3. confidence rating #$&*:ry ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The equation s * <0,3> + tu = v becomes s * <0, 3> + t * <-4, 3 > = < 2, -1 / 2 > or <0, 3s > + <-4 t, 3 t > = <2, -1/2 >. and finally <0 - 4 t, 3 s + 3 t > = < 2, -1/2 >. Since the two vectors are equal if an only if their two components are equal, this is equivalent to the two simultaneous equations -4 t = 2 3 s + 3 t = -1/2. The solution of these equations is t = -1/2, s = 1/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Let u = 4i - 3j, v = -3i + 4j , and w = 6i - 3j. Write the expression ||u|| ||v|| w in standard form. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt.(4^2 + (-3)^2) = 5, this the magnitude of both u and v. mag.u * mag.v = 25. 25w = 150i -75j. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: || u || = sqrt( 4^2 + 3^2) = 5 and || v || = sqrt(3^2 + 4^2) = 5 so that || u || || v || w = 5 * 5 * w = 25 * (6 i - 3 j ) = 150 i - 75 j. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Let u = 4i + j, v = 4i + 3j, w = -i + 2j. Find a vector of length 3 with the same direction as u - 2v + 2w. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To make typing this a bit easier, ‘A = u -2v + 2w. ‘A = -6i -j. ||’A|| = sqrt. (36 + 1) = 6.08. u’A = -6/6.08i - 1/6.08j. We multiply this by 3 to get a vector of magnitude 3 in the same direction, so we get -18/6.08i - 3/6.08j. confidence rating #$&*:y ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: u - 2v + 2w = -6 i - j so || u - 2 v + 2 w || = sqrt(37) and ( u - 2 v + 2 w ) / || u - 2 v + 2 w || = -6 sqrt(37) / 37 i - sqrt(37) / 37 j is a unit vector in the directio of u - 2 v + 2 w . A vector of magnitude 3 in this direction is therefore 3 ( -6 sqrt(37) / 37 * i - sqrt(37) / 37 * j ) = -18 sqrt(37) / 37 i - 3 sqrt(37) / 37 j &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I realize that putting a decimal in a fraction is not the neatest way to write these problems, but is it actually wrong? If so, I can easily use sqrt. And get the sqrt. Out of the denominator like you did. ------------------------------------------------ Self-critique rating:
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Given Solution: || v || = sqrt( cos^2(theta) + sin^2(theta) ) = sqrt(1) = 1. A vector of magnitude 1 is a unit vector. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!