1Q2)
Find the locus of the centroid of the tetrahedron of constant volume 64k^{3}, formed by the three co-ordinate planes and a variable plane.
The direction cosines of a variable line in two near-by positions are l, m, n; l+\delta l, m+\delta m, n+\delta n
Show that small angle \delta \theta between the two position is given by
(\delta \theta )^{2}=(\delta l)^{2}+(\delta m)^{2}+(\delta n)^{2}
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5 Answers
11l2 + m2 + n2 = 1
(l + δl)2 + (m + δm)2 + (n + δn)2 = 1
expanding we get
l2 + m2 + n2 + 2(lδl + mδm + nδn) + (δl)2 + (δm)2 + (δn)2 = 1
we know that
cosδθ = l(l + δl) + m(m + δm) + n(n +δn)
cosδθ = 1 + (lδl + mδm + nδn)
lδl + mδm + nδn = cosδθ - 1
substituting it
we get
1 + 2(cosδθ - 1) + (δl)2 + (δm)2 + (δn)2 = 1
(δl)2 + (δm)2 + (δn)2 = 2(1 - cosδθ)
(δl)2 + (δm)2 + (δn)2 = 4sin2δθ2
since δθ is very small si2δθ/2 = (δθ)2/4
we get the required result.
1Q2) let the equation of variable plane is lx + my + nz = p .....................(1)
also equations of co-ordinate planes are x = 0 , y = 0 , z = 0
solving for these four equations we getco-ordinates of vertices as
(0,0,0) ; ( 0,0,p/n) ; (p/l ,0,0) ; and (0,p/m,0)
let centroid be ( x1 , y1 , z1)
then x1 = 14(0+0+p/l+0) = p4l
similarly y1 = p4m , z1 = p4n
pl = 4x1 , pm = 4y1 , pn = 4z1 ..................(2)
also volume of tetrahedron = 64K3 ( given )
i.e., \frac{1}{6}\begin{vmatrix} p/l & 0& 0\\ 0& p/m& 0\\ 0& 0& p/n \end{vmatrix}=64K^{3}
or
p36lmn= 64K3
\frac{1}{6}\left(\frac{p}{}l \right)\left(\frac{p}{m} \right)\left(\frac{p}{n} \right)=64K^{3}
from (2) \frac{1}{6}\left( 4x_{1}\right)\left(4y_{1} \right)\left( 4z_{1}\right)=64K^{3} \Rightarrow x_{1}y_{1}z_{1}=6K^{3}
hence locus is
xyz = 6K3