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(19) Additional Questions

(19) Additional Questions. A= (3i-3j)m ,B=(i-4j)m, C =(-2i+5j) Magnitude and direction of the vector D=A+B+C Solution = (3+1-2)i+(-3-4+5)j D=(2i-2j) Dx=2 Dy=-2 thin Magnitude D= Dx ˆ2 +Dy ˆ2 D= 4+4 = 8

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(19) Additional Questions

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  1. (19) Additional Questions A= (3i-3j)m ,B=(i-4j)m, C =(-2i+5j) Magnitude and direction of the vector D=A+B+C Solution = (3+1-2)i+(-3-4+5)j D=(2i-2j) Dx=2 Dy=-2 thin Magnitude D= Dxˆ2+Dyˆ2 D= 4+4 = 8 Tan = Dx/Dy =2/-2 =-1 shift Tan =-45=315 (B) Magnitude and direction of the vector E=-A-B+C =(-3-1-2)i+(3+4+5)j -6i+12j Ex=-6, Ey=12 |E| = (-6)ˆ2+ 12 ˆ2 = 180 tan^-1 12/-6 =-63.4

  2. (20) A(6i-8j), B(-8i+3j), C(26i+19j) Determine a,b such that aA+bB+c=0 A(6i-8j)+b(-8i+3j)+(26i+19j) =0 (6ai-8aj)+(-8bi+3bj)+(26i+19j)=0 وبفصل الحدود المتشابهة نحصل على معادلتين 6a-8b+26=0 * 1= وبالقسمة على 6ai-8bi+26i=0 i -8aj+3bj+19j=0 j // // = -8a+3b+19=0 * 3 2 24a-32b+104=0,,,,,,,,,,, -24a+9b+57= -23b+161=0 b= 161/23=7 b=7وبالجمع On 1 6a-8b+26=0 =6a-8*7+26 =0 6a=30 a=30/6=5 a=5

  3. (21) R=2i+j+3k (a) Magnitude of the x,y,z Rx=2 Ry =1 Rz=3 (b) Magnitude of the R |R|= 2^2+1^2+3^2 = 14= 3.74 Angles between R xyz =tan^-1 R/Rx = tan^-1 3.74/2=61.8 =tan^-1 3.74/1=75.3 = tan^-1 3.74/3=51.26

  4. (22) V=2i+3j+k, ,w=4i+j+2k (a) Dot and Crosse product v.w=|V|W|cos |V|= (2ˆ2)+ ( 3 ˆ2)+1 = 14 |W|= (4ˆ2)+(1ˆ2 )+(2 ˆ2) = 21 v.w=(2i+3j+k).(4i+j+2k) = 8i.i+3j.j+2k.k =8*1+3*1+2*1 =13 v.w= 13 Angel cos =v.w/|v||w| =13/ 14 21 = 13/17.146 =0.7581 =cos^-1 0.7581 = 40.96

  5. 22 تكملة (b) vXw = + - + 2 3 1 4 1 2 =(3*2-1*1)i-(2*2-4*1)j+(2-3*4) =6i-0j-10k v.w= 6i+10k

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