Add the vectors 3i 5j 35 and 4i 2j 42 Find the magnitude

Add the vectors 3i + 5j = (3,5) and -4i + 2j = (-4,2). Find the magnitude of the answer and the angle in degrees that it makes with the positive x-axis. Round to the nearest tenth. b) Use the dot product (section 9.5) to find the angle between the vectors i + 3j = (1,3) and -2i + 4j = (- 2,4). Round this angle to the nearest hundredth of a degree.

Solution

2a) Let given a=3i+5j ,b=-4i+2j

c=a+b=-i+7j

Magnitude of given vectors |c|=sqrt(1+49)=sqrt(50)=7.071

Angle cos(Theta)=(a.b)/ (|a| |b|)=(-12+10)/{sqrt(9+25).sqrt(16+4)}=2/{sqrt(34)*sqrt(20)}

cos(Theta) ={2/{sqrt(34)*sqrt(20)}}=0.0766

Theta =cos^-1(0.0766)=1.494=1.5=1.6 since by rounding the nearest tength

Therefore

Magnitude of given vectors is 7.071

Angle of given two vectors is 1.6

b) Let The given vectors x=i+3j,y=-2i+4j

x.y=-2+12=10

|x|=sqrt(1+9)=sqrt(10)

|y|=sqrt(4+16)=sqrt(20)

cos (theta)={x.y}/{|x| |y| }=10/{sqrt(10) sqrt(20) } =0.7071

theta=cos^-1(0.7071) = 0.7854=0.800

Therefore

Angle =0.800 since nearest angle of hendreth