#$&* course Mth 277 6/15 11:45 am qa 09_1
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Given Solution: To move from P to Q we must move in the x direction from x = 3 to x = -4, a displacement of -7 units. So the `i component of the vector PQ is -7. Reasoning similarly the j and k components are 6 and -6, respectively, so PQ = -7 i + 6 j - 6 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer and your answer differ just by sign, and I'm not sure whether or not I'm right or not. ------------------------------------------------ Self-critique rating:
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Given Solution: The magnitude of the vector is found by the Pythagorean Theorem to be || PQ || = || -7 i + 6 j - 6 k || = sqrt( 7^2 + 6^2 + 6^2) = sqrt(121) = 11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003: What are the `i, `j and `k components of a unit vector in the direction of the vector PQ from the preceding two questions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From previous problem: The i vector is (3- -4) = 7. The j vector is (5-11) = -6. The k vector is (9-3) = 6. Divide each vector by the magnitude to get the length of each unit vector. `PQ = (7i - 6j + 6k)/11 `PQ = 0.636i - 0.545j + 0.545k Co nfidence rating: 3
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Given Solution: When you divide a vector by a positive number it becomes shorter by a factor equal to that number. For example if you divide a vector by 2, you end up with a vector of half the magnitude. A unit vector in the direction of a given vector is therefore obtained by dividing that vector by its magnitude. A unit vector in the direction of PQ is therefore u = PQ / || PQ || = (-7 i + 6 j - 6 k) / 11 = -7/11 i + 6/11 j - 6/11 k. A decimal approximation to this vector is roughly -.64 i + .55 j - .55 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004: What are the `i, `j and `k components of a vector parallel to PQ, having magnitude 20? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Take the unit vector we found that's pointing in the same direction as PQ. Since this new vector is too, the unit vector works for it as well. `RS = 0.636i - 0.545j + 0.545k Multiply it by 20 to get the value of the overall vector: `RS = 12.727i - 10.909j + 10.909k Confidence rating:
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Given Solution: We have found a unit vector in the direction of PQ. To get a vector of magnitude 20 in the same direction we just multiply this unit vector by 20. The vector 20 ( -7/11 i + 6/11 j - 6/11 k) = -140 / 11 i + 120 / 11 j - 120 / 11 k. A decimal approximation is -12.7 i + 10.9 j - 10.9 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Let P = (3, 5) and Q = (-1, 10), with `A the vector whose initial point is P and whose terminal point is Q. `q005. Let `v be the vector from the origin to point P. Sketch the points P and Q and the vectors `v and `A. Then sketch the points at the tip of each of the following vectors, provided the initial point of each is the origin: `v + .5 `A, `v + 1.5 `A, `v + 2.5 `A. Based on your sketch mark your estimated locations of the terminal points of each of the following, assuming the initial point for each to be the origin: `v + 2 `A `v + 3 `A `v - 1.5 `A. Estimate the coordinates of the terminal points of these vectors, based on your sketch. Calculate the coordinates of the points. How well can you fit a straight line to these points? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `v + 2 `A estimate: (-7, 25) calcualte: (-5, 15) `v + 3 `A estimate: (-10, 35) calculate: (-9, 20) `v + -1.5 `A estimate: (20, -5) calculate: (9, -2.5) These points all lie on the line through P and Q, since the vector `v will always point to P and the `A vector will always point to Q. Any multiplier in front of `A will only change the distance that the line is drawn. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. Is each of the following true or false, and why? 4 `i - 3 `j = 4 `i - 2 `j 3 x `i - 5 `j = 6 `i - 5 `j if, and only if, x = 2 || 4 `i - 3 `j || = || 3 `i + 4 `j || c * (4 `i - 3 `j) = (12 `i + 9 `j) if c = 3. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4 `i - 3 `j = 4 `i - 2 `j No, this isn't correct; if you draw out the vectors, their j component is different. 3 x `i - 5 `j = 6 `i - 5 `j if, and only if, x = 2 This is correct - if you assume that x = 2, then this expression evaluates to 6i - 5j = 6i - 5j; a reflexive equality. || 4 `i - 3 `j || = || 3 `i + 4 `j || While it doesn't look right, this is correct. The magnitude of both is calculated by applying the Pythagorean Theorem: (4i^2 + (-3j)^2) = mag1^2 mag1^2 = 25 mag1 = 5 The second is the exact same: (3i)^2 + (4j)^2 = mag2^2 mag2^2 = 25 mag2 = 5 c * (4 `i - 3 `j) = (12 `i + 9 `j) if c = 3. This is incorrect. When comparing two vectors, all components must be equal. When you distribute, the first vector is equal to 12i - 9j, compared to 12i + 9j. The magnitudes are equal, but the vectors themselves are not. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The equation 4 `i - 3 `j = 4 `i - 2 `j states that the `i and `j components are identical on both sides. This is not so since the `j components on the left is -3 and the `j component on the right is -2. The equation 3 x `i - 5 `j = 6 `i - 5 `j if, and only if, x = 2 is true if and only if 3x = 6 and -5 = -5. The latter is clearly true, and the former is true if and only if x = 2. The equation || 4 `i - 3 `j || = || 3 `i + 4 `j || states that sqrt( 4^2 + (-3)^2) = sqrt( 3^2 + 4^2). Bot expression are equal to sqrt(25) = 5, so the equation is true. If c = 3 the equation c * (4 `i - 3 `j) = (12 `i + 9 `j) becomes 3 * (4 `i - 3 `j) = (12 `i + 9 `j) . The left-hand side would have `j component -9 and the right-hand side would have `j component +9. This is clearly not so, so the equation would be false for c = 3. In fact there is no value of c which makes the equation true, since this would require that 4 c = 12 and -3 c = 9. These two equations yield different results for c. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Find the value(s) of c that make each of the following true. If no such value exists, explain why this is the case: c * ( 4 `i - 3 `j + 6 `k) = -48 `i + 36 `j - 72 `k c * || 4 `i + 3 `j - 6 `k || = || 6 `i - 4 `j + 3 `k || || 4 `i + 3 `j - 5 `k || = || 5 `i + 7 `j - c `k || YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: c * ( 4 `i - 3 `j + 6 `k) = -48 `i + 36 `j - 72 `k These two vectors are equal iff c = -12. c * || 4 `i + 3 `j - 6 `k || = || 6 `i - 4 `j + 3 `k || After applying the Pythagoren Theorem, we find that the magnitudes for each vector are equal to 61c^2 and 61, respectively. Therefore, for c = 1, the two vector's magnitudes are equal. || 4 `i + 3 `j - 5 `k || = || 5 `i + 7 `j - c `k || Since the equations for magnitude come out to 0 = 74 + c^2. In order to satisfy that, c would need to equal the sqrt(-74), which is impossible. There are no values for c that would satisfy this. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: c * ( 4 `i - 3 `j + 6 `k) = -48 `i + 36 `j - 72 `k implies that 4 c = -48, -3 c = 36 and 6 c = -72. Each of these three equations yields solution c = -12, so this value of c solves the equation. c * || 4 `i + 3 `j - 6 `k || = || 6 `i - 4 `j + 3 `k || becomes c * sqrt( 4^3 + 3^2 + 6^2) = sqrt(6^2 + 4^2 + 3^2), i.e., c * sqrt(61) = sqrt(61), with solution c = sqrt(61) / sqrt(61) = 1. || 4 `i + 3 `j - 5 `k || = || 5 `i + 7 `j - c `k || becomes sqrt(4^2 + 3^2 + 5^2) = sqrt(5^2 + 7^2 + c^2), i.e., sqrt(61) = sqrt(64 + c^2). This is true if and only if 61 = 64 + c^2, yielding c^2 = -3. Since the square of any real number is positive, there is no real-number solution to this equation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. If theta = pi / 6, then what is the magnitude of the vector sin(theta) * `i + cos(theta) * `j? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Applying the Pythagorean Theorem, we get the magnitude equal to sqrt( sin(pi/6)^2 + cos(pi/6)^2 ). sqrt( 0.5^2 + (sqrt(3)/2)^2 ) sqrt( 0.25 + 3/4 ) sqrt (1) 1 Therefore, the magnitude of this vector is 1 when theta is pi/6. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: sin(pi/6) `i + cos(pi/6) `j has magnitude || sin(pi/6) `i + cos(pi/6) `j || = sqrt( (sin^2(pi/6) + cos^2(pi/6) ). Since sin^2(theta) + cos^2(theta) = 1 for any value of theta, we get || sin(pi/6) `i + cos(pi/6) `j || = sqrt( (sin^2(pi/6) + cos^2(pi/6) ) = 1. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 3 ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!