Assignment 7 - Assignment 7 1 a u =(2,4 v =(1,3 We know...

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Assignment 71.a. ´u=(2,4´v=(1,3)We know that, cosθ´u.´v|´u||´v|cosθ=(2,4).(1,3)22+(4)212+(3)2¿(2) (1)+(4) (3)2010¿14200¿0.989θ=8.13°b. ´p=(1,4,5´q=(3,1,3)cosθ=(1,4,5).(3,1,3)(1)2+42+5232+(1)2+32¿(1) (3)+4(1)+(5) (3)4219¿8798
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¿0.283θ=73.55°2.6x3y+2=0Comparing to the general scalar equation in 2 space Ax+By+c=A=6, B=−3and C=Given a scalar equation in 2-space, the vector ´u=(A,B)is perpendicular to the line Ax+By+C=0and the slope of the Vector is BA.Hence vector ´u=(6,3)is perpendicular to the line and its slope is m36=123.´w=(4,1,3)+t(−2,1,7)To find another point, substitute any real number for t:If t=1,´w=(4,1,3)+1(2,1,7)=(4+1(2),1+(1) (1),3+1(7))=(2,0,10)Another way to write the equation of the line is using this new point (2,0,10´w=(2,0,10)+t(−2,1,7)Also using a scalar multiple of direction vector t(-2,1,7)´w=(2,0,10)+t(−4,2,14)
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4.a. ´a=(10,4,1´b=(3,2,2)´a.´b=(10,4,1).(3,2,2)¿10(3)+(4) (2)+1(2)¿40Since the dot product is positive the angle is Acute.b. ´p=(0,4,3´q=(7,2,1)´p.´q=(0,4,3).(7,2,1)¿0(7)+4(2)+(3) (1)¿11Since the dot product is negative the angle is Obtuse.5.(x, y)=(3,2)+t(2,4)Given the vector equation of this line, we know a point on the line is (3,2). We can also find the slope of the line using the direction vector. ´u=(A ,B), the slope of the line is mBASince the direction vector of this line is ´u=(3,2), the slope of the line is m42=2.Using the formula,y=m(xx1)+y
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y=2(x3)+2y=2x46.A direction vector parallel to the Y-axis is ´u=(0,1,0This vector can have any value for y, but the x and z coordinates must be 0.Also since a point on the line is P(-1,0,3)A vector equation is (x, y , z)=(−1,0,3)+t(0,1,0)7.´u=(4,1,1)´v=(1,3,1)´w=(2,5,13)´u.´v=(4,1,1).(1,3,1)¿4(1)+(1) (3)+1(1)¿0´v .´w=(1,3,1).(2,5,13)¿1(2)+3(5)+(1) (13)¿0´w .´u=(2,5,13).(4,1,1)¿2(4)+(5) (1)+(13) (1)¿0
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