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Three Dimensional Geometry

Short Answer Type

101.

Find the area of the triangle whose vertices are (1, 2, 4), (-2, 1, 2), (2, 4, -3).

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Long Answer Type

102. A line makes angle α, β, γ and δ with the diagonals of a cube, prove that

cos squared straight alpha space plus space cos squared straight beta space plus space cos squared straight gamma space space plus cos squared straight delta space equals space 4 over 3.

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103. If the edges of a rectangular parallelepiped are a, b, c, show that the angles between four diagonals are given by cos^–1

open parentheses fraction numerator straight a squared plus-or-minus straight b squared plus-or-minus straight c squared over denominator straight a squared plus straight b squared plus straight c squared end fraction close parentheses.

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104. Find the angle between two diagonals of a cube.

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105. Show that the line joining the middle points of two sides of a triangle is parallel to the third side and half of it in length.

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106. A variable line in two adjacent positions has direction cosines < l, m, n > and < l + δl, m + δm, n + δn >. Show that the small angle δθ between two positions is given by
(δθ )² = (δl)² + (δm)² + (δn)²

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107.
Find the angle between the two lines whose direction cosines are given by the equations:
l + m + n = 0, l² + m² – n² = 0

The given equation are
l + m + n = 0    ...(1)
and l² + m² – n² = 0    ....(2)
From (1), n = – (l + m)    ....(3)
From (2) and (3), we get,
l² + m² – (l + m)² = 0 or – 2 l m = 0
⇒ l m = 0
$therefore$ either l = 0
$therefore$ 1. l + 0 .m + 0. n = 0
Also, l+m+n = 0
Solving,
   $fraction numerator straight l over denominator 0 minus 0 end fraction space equals space fraction numerator straight m over denominator 0 minus 1 end fraction space equals space fraction numerator straight n over denominator 1 minus 0 end fraction$

$therefore$ $straight l over 0 space equals space fraction numerator straight m over denominator negative 1 end fraction space equals space straight n over 1$
or m = 0
$therefore$ 0.l + l.m + 0.n = 0
Also, l + m + n = 0
Solving,
$fraction numerator straight l over denominator 1 minus 0 end fraction space equals space fraction numerator straight m over denominator 0 minus 0 end fraction space equals space fraction numerator straight n over denominator 0 minus 1 end fraction$

$therefore$ $straight l over 1 space equals space straight m over 0 space equals space fraction numerator straight n over denominator negative 1 end fraction$
∴ direction ratios of the two lines are 0, – 1, 1 ; 1, 0, – 1
Let  θ be the angle between the lines
$therefore space space space space cos space straight theta space equals space fraction numerator left parenthesis 0 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis 0 right parenthesis space plus space left parenthesis 1 right parenthesis thin space left parenthesis negative 1 right parenthesis over denominator square root of left parenthesis 0 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared end root space square root of left parenthesis 1 right parenthesis squared plus left parenthesis 0 right parenthesis squared plus left parenthesis negative 1 right parenthesis squared end root end fraction space equals negative 1 half$
∴ acute angle θ between the lines is given by
$cos space straight theta space equals space 1 half space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space space space straight theta space equals space 60 degree$

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108. Find the angle between the two lines whose direction cosines are given by the equations:
2 l – m + 2 n = 0 and m n + n l + l m = 0

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109. Find the angle between the two lines whose direction cosines are given by the equations:
l + m + n = 0 and 2 l + 2 m – m n = 0

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110. Show that the straight lines whose direction cosines are given by the equations uI + vm + wn = 0, a I² + b m² + cn² = 0 are
(i) perpendicular if u² (b + c) + v² (c + a) + w² (a + b) = 0
(ii) parallel if