Question

if a variable line in two adjacent positions has direction cosines l,m,n and l+$l,m+$m,n+$n show that the small angle $x between two positions is given by ($x)^2=($l)^2+($m)^2+($n)^2.

Answer 1

Step 1:Let the direction cosines of the two lines be d1=(l,m,n)andd2=(l+δl,m+δm,n+δn)d1=(l,m,n)andd2=(l+δl,m+δm,n+δn)We know that angle θθ between two lines is given bycosθ=(d→1.d→2)cosθ=(d→1.d→2)It is given that the angle between the lines is δθδθ⇒cosδθ=(l,m,n).(l+δl,m+δm,n+δn)⇒cosδθ=(l,m,n).(l+δl,m+δm,n+δn)=l(l+δl)+m(m+δm)+n(n+δn)=l(l+δl)+m(m+δm)+n(n+δn)On expanding we get,⇒l2+m2+n2+lδl+mδm+nδn=0⇒l2+m2+n2+lδl+mδm+nδn=0Step 2:We know that l2+m2+n2=1l2+m2+n2=1Square of direction cosines is one.Therefore ⇒cosδθ=lδl+mδm+nδn+1⇒cosδθ=lδl+mδm+nδn+1..........(i)l2+m2+n2=1and(l+δl)2+(m+δm)2+(n+δn)2=1l2+m2+n2=1and(l+δl)2+(m+δm)2+(n+δn)2=1On expanding we get,⇒1+δl2+δm2+δn2+2lδl+2mδm+2nδn=1⇒1+δl2+δm2+δn2+2lδl+2mδm+2nδn=1⇒1+δl2+δm2+δn2+2(lδl+mδm+nδn)=1⇒1+δl2+δm2+δn2+2(lδl+mδm+nδn)=1Substituting for l2+m2+n2=1l2+m2+n2=1Step 3:1+δl2+δm2+δn2+2(lsl+msm+nδn)=11+δl2+δm2+δn2+2(lsl+msm+nδn)=1⇒δl2+δm2+δn2+2(cosδθ−1)=0⇒δl2+δm2+δn2+2(cosδθ−1)=0⇒δl2+δm2+δn2=2(1−cosδθ)⇒δl2+δm2+δn2=2(1−cos⁡δθ)But 1−cos2θ=2sin2θ/21−cos2θ=2sin2θ/2⇒δl2+δm2+δn2=2(2sin2δθ2)⇒δl2+δm2+δn2=2(2sin2δθ2)⇒δl2+δm2+δn2=4sin2δθ2⇒δl2+δm2+δn2=4sin2δθ2If θθ is small then sinθ≈θsinθ≈θ∴4δθ24∴4δθ24=δl2+δm2+δn2=δl2+δm2+δn2⇒δθ=δl2+δm2+δn2⇒δθ=δl2+δm2+δn2Hence proved
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