Question

In a meeting of mimes, mime 1 goes through a displacement d Overscript right-arrow EndScripts Subscript 1 Baseline equals left-parenthesis 2.32 m right-parenthesis i Overscript ̂ EndScripts plus left-parenthesis 6.85 m right-parenthesis j Overscript ̂ EndScripts and and mime 2 goes through a displacement d Overscript right-arrow EndScripts Subscript 2 Baseline equals left-parenthesis -7.48 m right-parenthesis i Overscript ̂ EndScripts plus left-parenthesis 9.9 m right-parenthesis j Overscript ̂ EndScripts. What are (a) vertical line d Overscript right-arrow EndScripts Subscript 1 Baseline times d Overscript right-arrow EndScripts Subscript 2 Baseline vertical line, (b) d Overscript right-arrow EndScripts Subscript 1 Baseline times d Overscript right-arrow EndScripts Subscript 2, (c) left-parenthesis d Overscript right-arrow EndScripts Subscript 1 Baseline plus d Overscript right-arrow EndScripts Subscript 2 Baseline right-parenthesis times d Overscript right-arrow EndScripts Subscript 2, and (d) the component of d Overscript right-arrow EndScripts Subscript 1 along the direction of d Overscript right-arrow EndScripts Subscript 2? Give your answers in standard SI units.

Answer 1

d1 = (2.32 m)i + (6.85 m)j

d2 = (-7.48 m)i + (9.9 m)j

a)

magnitude of

d1 x d2 = (9.9 m)*(2.32 m) - (-7.48 m)(6.85 m)

d1 x d2 = 74.206 m^2

b)

d1.d2 = (2.32 m)(-7.48 m) + (6.85 m)(9.9 m)

d1.d2 = 50.4614 m^2

c)

(d1+d2).d2 = [ (-5.16 m)i + (16.75 m)j ].[ (-7.48 m)i + (9.9 m)j ]

(d1+d2).d2 = 38.5968 + 165.825

(d1+d2).d2 = 204.4218 m^2

d)

The component of d1 along d2 is given by d1cos(theta)

Where theta is the angle between the vectors

cos(theta) = d1.d2/d1d2

cos(theta) = 50.4614 /(7.232*12.408)

cos(theta) = 0.5623

The component of d1 along d2 is given by 7.232*(0.5623) = 4.0668 m

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