#$&* course M 277 1-19 6Thanks a lot for recommending the Pre-Cal assignments. They were helpful on so many levels and brought lots and lots of stuff together!!!!! Especially the one on vectors and dot product. See my note at the end of the assignment if you have time. If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `aThe magnitudes are 10 and 8, and the angle between the vectors is the change in angle from 30 degrees to 90 degrees, or 60 degrees. The dot product is therefore dot product = product of magnitudes * cos(angle) = 10 * 8 * cos(60 deg) = 40. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. What are the x and y components of the vector v having magnitude 10 and angle 30 degrees, and vector w having magnitude 8 and angle 90 degrees? What do you get if you add the product of the two x components to the product of the two y components? How is this result related to the answer to the preceding exercise? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v x=10*cos(30) = 8.66, y=10*sin(30) =5 wx=8*cos(90) = 0, y=8*sin(90)=8 `sqrt(8.66^2 + 5^2) = approx. 10 `sqrt(0^2 + 8^2) = 8 where in previous answer we had 10*8, this are the same magnitudes found using Pythagorean Theorem. Which we mult by cos(90). confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe vector v has x component vx = 10 cos(30 deg) = 8.7, approx., and vy = 10 sin(30 deg) = 5. The vector w has x component wx = 8 cos(90 deg) = 0 and y component wy = 8 sin(90 deg) = 8. The product of the two x components is vx * wx = 8.7 * 0 = 0, and the product of the y components is vy * wy = 5 * 8 = 40. The sum of these products is 0 + 40 = 40, which is identical to the result of the preceding exercise in which we found the dot product of v and w. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??????I didn’t see the dot product relation but made another connection. Is this relevant or coincidental?????????????
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Given Solution: `aIf the preceding exercise generalizes then the dot product is the sum of the product of the x components and the product of the y components. In this case we would therefore say that the dot product is v1 * w1 + v2 * w2. The magnitudes of the two vectors are | v | = sqrt(v1^2 + v2^2) and | w | = sqrt(w1^2 + w2^2). The statement therefore says that v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I swear I didn’t look, even our variables are labled the same. Maybe I’ve had you instruct me for one or two courses to many, scary. Not quite there but next thing you know I will be able to find `sqrt(34,958,271,356.8473) in my head in a little less than 2.87sec.
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Given Solution: `aThe cosine of the angle theta between the vectors is found using the fact that dot product is product of magnitudes multiplies by the cosine of the angle, expressed in detail as v1 * w1 + v2 * w2 = sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2) * cos(theta). We easily rearrange the equation to get cos(theta) = [ v1 * w1 + v2 * w2 ] / [ sqrt(v1^2 + v2^2) * sqrt(w1^2 + w2^2)]. In this case we have cos(theta) = [ 2 * -7 + 3 * 4 ] / [ sqrt(2^3 + 3^2) * sqrt( (-7)^2 + 4^2) ] = -2 / ( sqrt(13) * sqrt(53) ] = -2 / 26.3 = .08 approx.. The angle is therefore the solution theta to the equation cos(theta) = .08, which gives us theta = 83 degrees, approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??????????????????????????????????????????????????????????????????????????????????????? I keep plugging these values into my calculator( I’m in degree mode) and I’m not getting cos(83deg) = .08, I’m getting .1217. cos(85) = .087, so is this where approx. comes in and does this corresponed to the `pi/3, because I solved the problem a totally different way and I’ve never been shown this material so is this all just dumb luck I keep coming up with answers that work or am I simply using another approach.????????????????????????????????????? I understand solution
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Given Solution: `aThe magnitude of a vector is found by taking the square root of the sum of the squares of its components, just as for a vector in 2-dimensional space. In this case we get magnitudes sqrt(2^2 + 4^2 + 5^2) = sqrt(45) = 6.7, approx., and sqrt((-3)^2 + 7^2 + 2^2) = sqrt(76) = 8.7 approx.. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That’s what I was thinking. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q006. What is the dot product of the vectors v = < 2, 4, 5 > and w = < -3, 7, 2 > ? What therefore is the angle between these vectors? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dot product is 2*-3 + 4*7 + 5*2 = 32 so I’m a little confused but I think cos(`theta) = 32/(6.7*8.7) = .549 approx `theta = cos^-1(.549) = approx. 56.7, I’m not positive this is correct but I believe so confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince dot product = product of magnitudes * cos(theta) we have cos(theta) = dot product / product of magnitudes. In the preceding problem we found the magnitudes of the vectors to be sqrt(45) and sqrt(76). So if we can find the dot product we can find the cosine of the angle and therefore the angle. The dot product is again the sum of the products of the invidual components, in this case 2 * (-3) + 4 * 7 + 5 * 2 = 32. Thus we have cos(theta) = 32 / ( sqrt(45) * sqrt(76) ) and theta = arccos[ 32 / (sqrt(45) * sqrt(76) ) = 56.8 deg, approx.. Note that these vectors can actually be constructed in 3-dimensional space, and if the construction is accurate the angle will be as indicated. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): So I used values you solved for in previous problem which I didn’t pay enough attention and were approx., so will using approx. instead of exact values( `sqrt(45) & `sqrt(76)) give you a good approx. answer or should exact values be used because it is is to throw off eqution???????? I guess really just an FYI question, not real important just curious.