qr code vb.net open source Let s call this Cartesian sum vector c = (xc,yc), so we have xc = 3 and in VS .NET

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The same thing happens with the other elements The y coordinate of a - b is ya yb, and the y coordinate of b - a is yb ya The rules of pre-algebra tell us that yb ya = ( ya yb) Therefore, we know that b - a = [ (xa xb), ( ya yb)] By definition, that s the Cartesian negative of a - b 4 We are given the two Cartesian vectors a = (4,5) and b = ( 2, 3) Their Cartesian sum is a + b = {[4 + ( 2)],[5 + ( 3)]} = (2,2) The individual Cartesian negatives are -a = ( 4, 5) and -b = (2,3) These vectors add up to a + ( b) = [( 4 + 2),( 5 + 3)] = ( 2, 2) In this case, the sum of the Cartesian negatives is equal to the negative of the Cartesian sum 5 Let s begin by working out a formula for the negative of a vector sum Suppose we re given two Cartesian vectors a = (xa,ya) and b = (xb,yb)
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The sum vector a + b is a + b = [(xa + xb),( ya + yb)] The negative of this sum vector is -(a + b) = [ (xa + xb), ( ya + yb)] Using the rules of pre-algebra, we can rewrite the right-hand side of this equation to get -(a + b) = [ xa + ( xb)],[ ya + ( yb)] Now let s go back to the original two vectors We can state their Cartesian negatives as -a = ( xa, ya) and -b = ( xb, yb) When we add these, we obtain -a + (-b) = [ xa + ( xb)],[ ya + ( yb)] That s the same thing we got when we worked out -(a + b), so we know that -(a + b) = a + (-b) 6 We are given the two polar vectors a = (p /2,4) and b = (p,3) We want to find their polar sum First, we convert the vectors to Cartesian form When we do that, we get a = {[4 cos (p /2)],[4 sin (p /2)]} = [(4 0),(4 1)] = (0,4) and b = [(3 cos p),(3 sin p)] = {[(3 ( 1)],(3 0)]} = ( 3,0) When we add these, we obtain a + b = {[0 + ( 3)],(4 + 0)} = ( 3,4)
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Let s call this Cartesian sum vector c = (xc,yc), so we have xc = 3 and
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yc = 4
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The point defined by these coordinates lies in the second quadrant of the Cartesian plane We want to know the polar sum vector c = (qc,rc), where qc is the direction angle of c and rc is the magnitude of c Using the applicable angle-conversion formula, we get qc = p + Arctan [4/( 3)] = p + Arctan ( 4/3) That s an irrational number If we want to be exact, we must leave it in this form; there s no way to make it simpler! A calculator set to work in radians can give us approximate values to four decimal places of Arctan ( 4/3) 09273 and p 31416 From this, we can calculate qc 22143 Using the formula for the polar magnitude, we obtain rc = (xc2 + yc2)1/2 = [( 3)2 + 42]1/2 = (9 + 16)1/2 = 251/2 = 5 This value is exact Putting the coordinates into an ordered pair, we derive our exact final answer as c = a + b = (qc,rc) = {[p + Arctan ( 4/3)],5} The approximate-angle version is c = a + b (22143,5) Don t get confused here This ordered pair looks deceptively like the rendition of a vector in the Cartesian plane, but it really defines the vector in the polar coordinate plane The first coordinate is in radians, and the second coordinate is in linear units
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7 To find the polar negative of the vector we derived in the solution to Problem 6, we reverse the direction but leave the magnitude the same In this situation, 0 qc < p, so we should add p to the angle to reverse the direction That gives us the exact answer as -(a + b) = {[p + p + Arctan ( 4/3)],5} = {[2p + Arctan ( 4/3)],5} If we say that 2p 62832, then we can approximate the angle to four decimal places and define the vector as -(a + b) (53559,5) 8 The original two vectors are a = (p /2,4) and b = (p,3) To find the polar negatives, we reverse the directions but leave the magnitudes the same We want to keep the angles less than 2p without letting either of them become negative In this case, that means we should add p to qa, but we should subtract p from qb When we make these changes, we get -a = (3p /2,4) and -b = (0,3) We must be careful to avoid confusion about what the coordinates of b actually mean The first entry in the ordered pair is an angle, while the second entry is a radius 9 This time, we want to find the polar sum of the vectors -a = (3p /2,4) and -b = (0,3) Converting them to Cartesian form, we get -a = {[4 cos (3p /2)],[4 sin (3p /2)]} = {(4 0),[ 4 ( 1)]} = (0, 4)
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