#$&* course Mth 277 qa 09_1
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Given Solution: To move from P to Q we must move in the x direction from x = 3 to x = -4, a displacement of -7 units. So the `i component of the vector PQ is -7. Reasoning similarly the j and k components are 6 and -6, respectively, so PQ = -7 i + 6 j - 6 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002: What is the magnitude of the vector defined in the preceding problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The magnitude of the vector is found by squaring each component and adding them together. Then, you take the squareroot of that quantity. sqrt((-7)^2 + 6^2 + (-6)^2)=sqrt(49+36+36)=sqrt(121)=11 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: As stated earlier, to find the unit vector you should take the PQ vector and divide by the magnituded of that vector So, (-7/11)i+(6/11)j+(-6/11)k Approx. -.636i+.545j-.636k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When you divide a vector by a positive number it becomes shorter by a factor equal to that number. For example if you divide a vector by 2, you end up with a vector of half the magnitude. A unit vector in the direction of a given vector is therefore obtained by dividing that vector by its magnitude. A unit vector in the direction of PQ is therefore u = PQ / || PQ || = (-7 i + 6 j - 6 k) / 11 = -7/11 i + 6/11 j - 6/11 k. A decimal approximation to this vector is roughly -.64 i + .55 j - .55 k. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q004: What are the `i, `j and `k components of a vector parallel to PQ, having magnitude 20? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since we have found the vector that is parallel to PQ, we just need to multiply that vector by 20. 20(-7/11i+6/11j-6/11k)=-12.7i+10.9j-10.9k confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: 3+2(-4)=-5i 5+2(5)=15j (-5,15) 3+3(-4)=-9i 5+3(5)=20j (-9,20) 3-1.5(-4)=9i 5-1.5(5)=-2.5j (9,-2.5) If I did what was asked then you can create a decently straight line through the points. Although, the points are not in the order given to make them line up properly.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not seem to grasp this question as easily as the others. Finding the vectors was easy and was stated very clear. However, the rest left me slightly unsure. ------------------------------------------------ Self-critique rating:OK ********************************************* 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: The first question is obviously false because the j components are obviously different. The matching components on both sides of the equal signs need to be identical. The second question is true because with x being two, the components on both sides are equal to eachother. 3(2)i-5j=6i-5j. Obviously, the only number that can be substituted into x is 2 for the equation to work properly. Since both sides of the equation ask for the magnitude, the signs do not matter. This is because each number is squared so the sign, whether negative or positive, disappears. The last equation is false because the signs for the j component do not match up. After distributing the c value on the left side, the components become 12i-9j which does not equal 12i+9j. 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):OK ------------------------------------------------ Self-critique rating:OK ********************************************* 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: The value of c for the first equation is -12. This is shown by what number should be distributed to make the components on the left and right equal. 4(-12)=-48;-3(-12)=36;6(-12)=-72. To find out the value of c, you need to take the magnitude of the equations on both side. The left=c*sqrt(61) and the right=sqrt(61).Therefore, the value of c would be 1. There is no value for c that can make these two equations equal. This is shown because the first magnitude is less than the magnitude of the first two components on the left. Since we are squaring the numbers, there is no way to decrease the magnitude with a number so there is no value for c that would make the equations equal. 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): I could have better explained where my numbers came from and how I achieved my answers. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q008. If theta = pi / 6, then what is the magnitude of the vector sin(theta) * `i + cos(theta) * `j? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using the unit circle, we know that the sin(theta)=.5 and the cos(theta)=(sqrt(3))/2. When doing the magnitude, we get sqrt((.5)^2+(.866)^2)=1 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):OK ------------------------------------------------ Self-critique rating:OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!