#$&* course Mth 277 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: The distance between the point (x, y, z) and the point (3, -4, 2) is sqrt((3 - x)^2 + (-4 - y)^2 + (2 - z)^2) = sqrt( (x - 3)^2 + (y + 4)^2 + (z - 2)^2). This distance is 5 if and only if sqrt( (x - 3)^2 + (y + 4)^2 + (z - 2)^2). = 5. Squaring both sides we obtain (x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. Using the equation from the preceding, find the value of y if we know that x = 2 and z = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If x=2 and z=1 then: (y+4)^2+2=25, y^2+8y+18=25, 0=y^2+8y-7. Using the quadratic formula, y=-8.795831523 or 0.795831523 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ If x = 2 and z = 1 then our equation (x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25 becomes (2 - 3)^2 + (y + 4)^2 + (1 - 2)^2) = 25, or 1 + (y+4)^2 + 1 = 25 so that (y + 4)^2 = 23 and (y + 4) = +- sqrt(23). Solving for y we get y = sqrt(23) - 4 or y = -sqrt(23) - 4.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q003. Using the equation from the first question, substitute z = 0. The resulting equation describes a circle. What are its center and radius? Answer the same questions if you substitute z = 1 rather than z = 0. Answer the same questions if you substitute z = -1 rather than z = 0. Answer the same questions if you substitute z = -4 rather than z = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If z=0, then the equation is 21=(x-3)^2+(y+4)^2 with center (3,-4) and radius sqrt(21) If z=1, then 24=(x-3)^2+(y+4)^2 with center (3,-4) and radius sqrt(24) If z=-1 then 16=(x-3)^2+(y+4)^2 with center (3,-4) and radius 4. If z=-4 then -11=(x-3)^2+(y+4)^2, which does not exist because the radius must be imaginary. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Our equation is (x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25. If z = 0 then our equation becomes (x - 3)^2 + (y + 4)^2 + (0 - 2)^2) = 25, which we easily rearrange to get (x - 3)^2 + (y + 4)^2 = 21. This is the equation of a circle in the x-y plane centered at (3, -4) and having radius sqrt(21), which is roughly 4.6. If z = 1 then (z - 2)^2 becomes 1 and our equation becomes (x - 3)^2 + (y + 4)^2 = 24. This is the equation of a circle in the x-y plane centered at (3, -4) and having radius sqrt(24), which is roughly 4.9. If z = -1 we get a circle of radius 6, centered at (3, -4). If z = -4 we get the equation (x-3)^2 + (y+4)^2 = -11. Since squares can't be negative, there is no real-number solution for (x, y), and there will be no z = -4 point on this graph. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If z=-1, then the equation must be (x-3)^2+(y+4)^2+(-1-2)^2=25, so (x-3)^2+(y+4)^2+(-3)^2=25, (x-3)^2+(y+4)^2+9=25, then (x-3)^2+(y+4)^2= 16, so the radius must be 4. I do not understand why you say that the radius is 6.
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Given Solution: The equation (x - 3)^2 + (y + 4)^2 + (z - 2)^2) = 25 becomes x^2 - 6 x + 9 + y^2 + 8 y + 16 + z^2 - 4 z + 4 = 25 which in standard form is x^2 - 6 x + y^2 + 8 y + z^2 - 4 z + 4 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q005. Is the vector 8 `i - 4 `j + 5 `k a multiple of the vector 4 `i + 2 `j - 5/2 `k? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8i-4j+5k is a multiple of 4i+2j-5/2k if 8i-4j+5k= c(4i+2j-5/2k) where c is a constant. For the j and k components to match, c must be -2. For the I components to match c must be 2. Therefore, because c can only be one number, the vector 8i-4j+5k is not a multiple of 4i+2j-5/2k. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The vector 8 `i - 4 `j + 5 `k is a multiple of the vector 4 `i + 2 `j - 5/2 `k if, and only if, there exists a constant c such that 8 `i - 4 `j + 5 `k = c ( 4 `i + 2 `j - 5/2 `k ). Setting the `i, `j and `k components of the left-hand side respectively equal to the same components of the right-hand side we get the three equations 8 = 4 c -4 = 2 c 5 = -5/2 c The first yields solution c = 2. The last two yield solution c = -2. c can't take both values at the same time, so the equations are not simultaneously true. It follows that neither vector is a multiple of the other. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q006. How does your answer to the preceding determine whether or not the two vectors are parallel? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If one vector is a multiple or factor of another, the two are parallel. They are parallel because the direction is the same, only the magnitude is changed by the scalar proportion. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are parallel if, and only if, one is a multiple of the other. So the solution to the preceding shows that neither vector is a multiple of the other. STUDENT SOLUTION (not quite complete): If a vector is a multiple of another, the two vectors are parallel. Instructor Response: Right idea but the statement you base the conclusion on needs to be stronger. Had one vector been a multiple of the other, then your statement would allow us to conclude that the vectors are parallel. However your statement doesn't address what happens if neither vector is a multiple of the other. That would require a statement equivalent to the following: If two vectors are parallel, then each is a scalar multiple of the other. We can combine the two statement into the single statement Two vectors are scalar multiples of one another if and only if they are parallel. This statement is equivalent to the statement in the given solution. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q007. What are the lengths of the sides of the triangle whose vertices are (4, 3 -2), (5, -1, 3) and (6, 4, 1)? Sketch a triangle whose sides have these lengths, as best you can in a few minutes without meticulously measuring everything (i.e., try to keep the sides in approximately the right proportion). Based on your sketch does it seem plausible that this is a right triangle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the point (4,3,-2) to the point (5,-1,3) the length is sqrt(1^2+(-4)^2+5^2))= sqrt(42). From the point (5,-1,3) to (6,4,1) the length is sqrt(1^2+5^2+(-2)^2)= sqrt(30). From the point (6,4,1) to (4,3,-2) the length is sqrt((-2)^2+(-1)^2+(-3)^2)= sqrt(14). This is a right triangle if: (sqrt(30))^2 +(sqrt(14))^2= (sqrt(42))^2, but it does not so this is not a right triangle. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!