3D

If direction cosines of a variable line in 2 adjacent points be l,m,n and l+δl,m+δm,n+δn .Show that small angle δθ between two positions is given by δθ2=δl2+δm2+δn2

6 Answers

1
Che ·

cos δθ=l.(l+δl) +m.(m+δm) +n.(n+δn)

=l2+m2+n2+l.δl+m.δm+n.δn

=1+l.δl+m.δm+n.δn

cos δθ-1= l.δl+m.δm+n.δn

1-cos δθ=-(l.δl+m.δm+n.δn)

we know 1-cos δθ=2sin2 δθ/2 .........for small δθ.....1-cos δθ=2(δθ/2)2

(δθ)2/2= -(l.δl+m.δm+n.δn)...............................................i

now we know that l2+m2+n2=1......................ii

(l+δl)2+(m+δm)2+(n+δn)2=1..........................iii

now just subtract eq 3 from 2 and substitute 1.......u ill get teh req ans

11
Devil ·

Hint :- Expansion of cosx.

I cud not get much from joker's soln.

1
Che ·

l2+m2+n2+2(l.δl+m.δm+n.δn)+(δl)2+(δm)2+(δn)2 =1.................iii

l2+m2+n2=1........................ii

iii-ii

2(l.δl+m.δm+n.δn)+(δl)2+(δm)2+(δn)2 =0

from i

2[-(δθ)2/2] + (δl)2+(δm)2+(δn)2=0

(δθ)2= (δl)2+(δm)2+(δn)2

hp.....

1
Che ·

btw wat u din get soumik in my soln [2]

11
Devil ·

No its ok.....I used expansions.

24
eureka123 ·

tell ur method also soumik..

and thx thejoker[1]

Your Answer

Close [X]