#$&* course Mth 277
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Solution is not given, but I fell confident in my answer. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Let u = <-4,3> and v = <2,-1/2>. Find scalars s and t so that s * <0,3> + tu = v. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are given that s*<0,3>+t*<-4,3>=<2,-0.5> So lets solve for the xcomponent to find the value of t. s*0+t*(-4)=2 -4t=2 t=-0.5 Now lets solve for y values and plug in the value we solved for t. 3s+(-0.5)(3)=-.5 3s=1 s=1/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): We solved this question using different methods, but I believe my solution to be suffiecient. ------------------------------------------------ Self-critique rating: ********************************************* Question: Let u = 4i - 3j, v = -3i + 4j , and w = 6i - 3j. Write the expression ||u|| ||v|| w in standard form. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, we need to find the magnitudes of u and v. ||u||=sqrt((4)^2+(-3)^2)=sqrt(25)=5 ||v||=sqrt((-3)^2+(4)^2)=sqrt(25)=5 Given the expression ||u|| ||v|| w (5)(5)(6i-3j)=150i-75j confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: || u || = sqrt( 4^2 + 3^2) = 5 and || v || = sqrt(3^2 + 4^2) = 5 so that || u || || v || w = 5 * 5 * w = 25 * (6 i - 3 j ) = 150 i - 75 j. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Our answers are different,but after looking at your solution, I understand my mistake. ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Let u = 4i + j, v = 4i + 3j, w = -i + 2j. Find a vector of length 3 with the same direction as u - 2v + 2w. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u-2v+2w (4i+j)-2(4i+3j)+2(-i+2j)=-6i-j magnitude=sqrt((-6)^2+(-1)^2)=sqrt(37) So the unit vector would be 6/(sqrt(37))-1/(sqrt(37)) since it has a length of 3, the answer is 3(6/(sqrt(37))-1/(sqrt(37)))=18/sqrt(37)i-3/sqrt(37)j confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Since the unit vector is v / || v || || v ||=sqrt(cos^2(theta)+sin^2(theta)) trig identity cos^2(theta)+sin^2(theta)=1 sqrt(1)=1 So, the unit vector is v/1. Therefore, the vector v is a unit vector for any angle given. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!