#$&* course Mth 277 Question: Find u + v, u - v, (5/2)u, and 2u + 3v for the following vectors: u = <1,2,-3>, v = < -1,-2,3>.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): No solution given, but I believe my answer to be correct. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: Find the standard form equation of the sphere with center (-1,2,4) and radius 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: sqrt((x+1)^2+(y-2)^2+(z-4)^2)=2 (x+1)^2+(y-2)^2+(z-4)^2=4 x^2+2x+1+y^2-4y+4+z^2-8z+16=4 x^2+2x+y^2-4y+z^2-8z+17=0 is the standard form confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A point (x, y, z) is on the given sphere if its distance from (-1, 2, 4) is 2, so that sqrt( (x - (-1))^2 + (y - 2)^2 + (z - 4)^2 ) = 2 and (x + 1)^2 + (y - 2)^2 + (z - 4)^2 = 4. This is the equation of the sphere in one form. Expanding the squares we obtain x^2 + 2 x + 1 + y^2 - 4 y + 4 + z^2 - 8 x + 16 = 4 which we rearrange to the standard form x^2 + 2 x + y^2 - 4 y + z^2 - 8 z + 13 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I got 17 instead of the 13 seen in your solution. I reworked my problem several times and still get +17. Maybe you can see where I made my mistake. ------------------------------------------------ Self-critique rating:3
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Given Solution: Completing the squares we obtain (x^2 - 2 x + 1 - 1) + (y^2 - 6 y + 9 - 9) + (z^2 + 12 z + 36 - 36) = 17 which can be written as (x - 1)^2 - 1 + (y - 3)^2 - 9 + (z + 6)^2 - 36 = 17 and finally as (x - 1)^2 + (y - 3)^2 + (z + 6)^2 = 63 This sphere is centered at (1, 3, -6) and has radius sqrt(63) = 3 sqrt(7). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the standard representation and length of PQ when P = (-3,1,4) and Q = (2,-4,-3). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2-(-3)=5 -4-1=-5 -3-4=-7 5i-5j-7k sqrt((5)^2+(-5)^2+(-7)^2)=sqrt(99) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: PQ = (2 - (-3) ) i + (-4 - 1) j + (-3 - 4) k = 5 i - 5 j - 7 k. || PQ || = sqrt( 5^2 + 5^2 + 7^2) = sqrt(99). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find a unit vector in the direction of v = <-1, sqrt(3), 4>. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v=<-1,sqrt(3),4> magnitude of v=sqrt((-1)^2+(sqrt(3))^2+(4)^2)=sqrt(1+3+16)=sqrt(20) unit vector=<-1/sqrt(20),sqrt(3)/sqrt(20),4/sqrt(20)> confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: || v || = sqrt( 1^2 + sqrt(3) ^ 2 + 4^2 ) = sqrt( 26 ) so a unit vector in the direction of v is v / || v ||= < -1, sqrt(3), 4 > / sqrt(26) = <-sqrt(26) / 26, sqrt(78) / 26, 4 sqrt(26) / 26)> . 4 sqrt(26) / 26 is 2 sqrt(26) / 13. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am somewhat confused what you did. First, I do not know how you got sqrt(26) for the magnitude. If I am correct, I think you squared 3 rather than squaring sqrt(3). Then, I thought the unit vector was the vector divided by the magnitude so your other work confuses me. ------------------------------------------------ Self-critique rating:OK
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Given Solution: In the x-y plane y = cos(x) consists of a sinusoidal function oscillating between the lines y = -1 and y = 1, with period 2 pi radians, and containing the point (0, 1). The surface in 3 dimensions repeats this same curve for every value of z, so that the graph represents a wavy curtain hanging vertically downward, intersecting the xy plane along the sinusoidal curve. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Determine if u = 2i + 3j + -4k is parallel to v = <1,-3/2,2>. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Yes, u and v are not parallel to eachother. This can easily be determined without much work. Since <2,3,-4> is not a scalar multiple of <1,-1.5,2> these are not parallel to eachother. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are parallel if the angle between them is 0 or pi radians (180 degrees), meaning that the cosine of the angle between them is 1 or -1. u dot v = || u || || v || cos(theta) so that cos(theta) = u dot v / (|| u || || v || ) = (2 * 1 + 3 * (-3/2) + (-4 * 2) ) / ( sqrt(2^2 + 3^2 + 4^2) * sqrt( 1^2 + (3/2)^2 + 2^2) ) = (-21/2) / (sqrt( 29) sqrt(29/4). This is not 1 or -1, so the cosine is neither 0 nor pi rad (i.e., 180 deg). The vectors are therefore not parallel. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Can you not figure this out using scalar multiples? ------------------------------------------------ Self-critique rating:3
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Given Solution: The sides can be represented by the vectors AB = < 4, 1, 4 >, BC = < -2, 3, 0 > and AC = < 2, 4, 4 >. The magnitudes of these vectors are respectively sqrt(33) sqrt(13) sqrt(36). None of the sides are the same length so the triangle is not isosceles. The sum of the squares of the shorter two side is 33 + 13 = 46, which is not equal to the sum of the longest, so the triangle is not a right triangle. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): After seeing the sloution, it makes so much sense how to find if it is a right triangle and should have been able to solve it without help. ------------------------------------------------ Self-critique rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!