#$&* course Mth 277 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: v dot w = 4 * 3 + 1 * 2 = 12 + 2 = 14. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V dot w= 40+50-10= 80, not orthogonal If the vectors were orthogonal then v dot w would equal zero. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are orthogonal if the angle between them is 90 deg, i.e., if and onlye if their dot product is zero. The dot product of these vectors is 5 * 8 - 5 * (-8) + 5 * (-2) = 40 + 40 - 10 = 70. They are not orthogonal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the angle between v = 2i +3 k and w = -j + 4k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Cos(theta)= (a dot b)/(magnitude of a by the magnitude of b). Cos(theta)= (-2+12)/(sqrt(13)*sqrt(17))= 10/sqrt(221)= 0.6726 Theta= inverse cosine (0.67267)= 47.726 degrees or about 4pi/15 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since v dot w = || v || || w || cos(theta) we have theta = cos^-1 ( v dot w ) || v || || w || = cos^-1 ( 10 / (sqrt(13) * sqrt( 17) ) = cos^-1 (.67) = 48 degrees, approx., or roughly.8 radians. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find two distinct unit vectors orthogonal to both v = i + 2j -2k and w = i + j - 2k. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ai+bj+ck must be orthogonal to both vectors v and w, so a+2b-2c=0 and a+b-2c=0 must exist simultaneously. Therefore b=0 and a=2c and the vector 2ci+ck is orthogonal to v and w. This vector has magnitude of sqrt((2c)^2+c^2)= sqrt(5c^2)= (plus or minus) c*sqrt(5). And the unit vector that is orthogonal to vectors v and w is c^2*2*sqrt(5)i +c^2*sqrt(5)k, or -c^2*2*sqrt(5)i -c^2*sqrt(5)k. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Suppose a i + b j + c k is orthogonal to both. Then the dot product of this vector with each of the given vectors is zero, and we have a + 2 b - 2 c = 0 a + b - 2 c = 0 Subtracting the second equation from the first we get b = 0. With this value of b both our first and our second equation become a - 2 c = 0 so that a = 2 c. Any vector of the form 2c i + c k is therefore orthogonal to our two vectors. Any such vector has magnitude sqrt( (2 c)^2 + c^2) = sqrt( 5 c^2) = sqrt(5) | c |. If c is positive then | c | = c and our vector is (2 c i + c k ) / (sqrt(5) c) = 2 sqrt(5) / 5 i + sqrt(5) / 5 k. If c is negative then | c | = - c and our vector will be (2 c i + c k ) / (- sqrt(5) c) = - 2 sqrt(5) / 5 i - sqrt(5) / 5 k. Our two solution vectors are equal and opposite. Each is a unit vector perpendicular to the two given vectors. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Let v = i - j + 4k and w = -i + 3j + 2k. Find cos(theta). Find s such that v is orthogonal to sv - w. Also find t such that v - tw is orthogonal to w YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Cos(theta)= (-1-3+8)/(sqrt(18)*sqrt(14))= 4/sqrt(252). If v=i-j+4k and sv-w= si-sj+4sk+i-3j-2k. The dot product of v and sv-w= si^2+i^2+sj^2+3j^2+16sk^2-8k^2, which must equal zero for v and sv-w to be orthogonal. Since i^2+j^2+k^2=1, then the equation is s+1+s+3+16s-8=0, then 18s-4=0, then 18s=4, so s=4/18. v-tw= i-j+4k+ti-3tj-2tk. The dot product of v-tw and w is -i(i+ti)+3j(-j-3tj)+2k(4k-2tk), which equals zero if orthogonal. So, i^2(-1-t)+j^2(-3-9t)+k^2(8-4t)=0. Then, -14t+4=0, so t=4/14. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: cos(theta) = v dot w / ( || v || || w ||) = 4 / (sqrt(18) sqrt(14) ) = 4 / (12 sqrt(7) ). The condition v orthogonal to s v - w is v dot (s v - w ) = 0 (i - j + 4 k ) dot ( (s - 1) i + (-s + 3) j + (4 s + 2) k ) = 0 which becomes s - 1 + s - 3 + 16 s + 8 = 0 so that 18 s = 4 and s = 4 / 18. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find the work performed when a force F = (6/11)i - (2/11)j + (6/11)k is applied to an object moving along the line from P(3,5,-4) to Q(-4,-9,-11). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: W=F*D D=PQ= <-7, -14, -7> W= (6/11 *-7) + (-2/11 * -14) + (6/11 * -7) = 28/11 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The work is F dot `ds = ( (6/11)i - (2 / 11) j + (6 / 11) k ) dot (-7 i - 14 j - 7 k ) = -42/11 + 28 / 11 - 42 /11 = 28 / 11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Mth 277 If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: v dot w = 4 * 3 + 1 * 2 = 12 + 2 = 14. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Determine whether v = 5i - 5j + 5k and w = 8i - 10j -2k are orthogonal. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: V dot w= 40+50-10= 80, not orthogonal If the vectors were orthogonal then v dot w would equal zero. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Two vectors are orthogonal if the angle between them is 90 deg, i.e., if and onlye if their dot product is zero. The dot product of these vectors is 5 * 8 - 5 * (-8) + 5 * (-2) = 40 + 40 - 10 = 70. They are not orthogonal. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK