Question

A displacement B of 100m from the origin at the angle of 37° above the x-axis is the result of three successive displacement b1 of 100m along a negative x- axis and displacement b2 of 200m at the angle of 150°above the x- axis and displacement b2 , find b3

Answer 1

A displacement B of 100m from the origin at the angle of 37° above the x-axis is the result of three successive displacement b1 of 100m along a negative x- axis and displacement b2 of 200m at the angle of 150°above the x- axis and displacement b2.

so, B = 100 cos37° i + 100 sin37° j

= 80 i + 60 j

b1 = -100 i

b2 = 200cos150° i + 200sin150° j

= -100√3 i + 100 j

here, B = b1 + b2 + b3

⇒80i + 60 j = -100i - 100√3 i + 100j + b3

⇒80i + 60j = -100(1 + √3)i + 100j + b3

⇒b3 = (180 + 100√3)i - 40j

magnitude of b3 = √{(180 + 100√3)² + (-40)²} ≈ 355m

angle made by b3 =360° - tan-¹[ 40/(180 + 100√3)]

= 360 - 6.46 = 353.54°

Answer 2

$\huge\bold\purple{Answer:-}$

B = 100 cos37° i + 100 sin37° j

= 80 i + 60 j

b1 = -100 i

b2 = 200cos150° i + 200sin150° j

= -100√3 i + 100 j

here, B = b1 + b2 + b3

⇒80i + 60 j = -100i - 100√3 i + 100j + b3

⇒80i + 60j = -100(1 + √3)i + 100j + b3

⇒b3 = (180 + 100√3)i - 40j

magnitude of b3 = √{(180 + 100√3)² + (-40)²} ≈ 355m

angle made by b3 =360° - tan-¹[ 40/(180 + 100√3)]

= 360 - 6.46 = 353.54°