#$&* course Mth 277 9/9/12 11pm If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Let u = <-4,3> and v = <2,-1/2>. Find scalars s and t so that s * <0,3> + tu = v. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: s3j -4t i +3tj = 2i - .5j. -4t i = 2i. t = -.5 3s +3t = -.5 3s + 3(-.5) = -.5. 3s = 1. s= 1/3. t= -1/2 s=1/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The equation s * <0,3> + tu = v becomes s * <0, 3> + t * <-4, 3 > = < 2, -1 / 2 > or <0, 3s > + <-4 t, 3 t > = <2, -1/2 >. and finally <0 - 4 t, 3 s + 3 t > = < 2, -1/2 >. Since the two vectors are equal if an only if their two components are equal, this is equivalent to the two simultaneous equations -4 t = 2 3 s + 3 t = -1/2. The solution of these equations is t = -1/2, s = 1/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Let u = 4i - 3j, v = -3i + 4j , and w = 6i - 3j. Write the expression ||u|| ||v|| w in standard form. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ||4i - 3j||. sqrt( 4^2 + 3^2) = 5. ||-3i + 4j|| = 5. 25* (6i - 3j) 150 i - 75j. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: || u || = sqrt( 4^2 + 3^2) = 5 and || v || = sqrt(3^2 + 4^2) = 5 so that || u || || v || w = 5 * 5 * w = 25 * (6 i - 3 j ) = 150 i - 75 j. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Let u = 4i + j, v = 4i + 3j, w = -i + 2j. Find a vector of length 3 with the same direction as u - 2v + 2w. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = 4i + j -2v= -2(4i + 3j) = -8i -6j. 2w = 2*(-i + 2j) = -2i +4j. u - 2v + 2w = -6i -j. 3*(-6i -j)/||-6i -j||. -18/sqrt(37) i -3/sqrt(37) j. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: u - 2v + 2w = -6 i - j so || u - 2 v + 2 w || = sqrt(37) and ( u - 2 v + 2 w ) / || u - 2 v + 2 w || = -6 sqrt(37) / 37 i - sqrt(37) / 37 j is a unit vector in the directio of u - 2 v + 2 w . A vector of magnitude 3 in this direction is therefore 3 ( -6 sqrt(37) / 37 * i - sqrt(37) / 37 * j ) = -18 sqrt(37) / 37 i - 3 sqrt(37) / 37 j &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: Show that the vector v = cos(theta)i + sin(theta)j is a unit vector for any angle theta. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Magnitude of this is sqrt of the components squared which is always 1 in the case of sin and cos theta. Thus always a unit vector. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: || v || = sqrt( cos^2(theta) + sin^2(theta) ) = sqrt(1) = 1. A vector of magnitude 1 is a unit vector. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ________________________________________ #$&*