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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

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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

The given equations are
l + m + n = 0    ....(1)
and 2 l + 2 m – m n = 0    ...(2)
From (1), l = – (m + n)    ...(3)
From (2) and (3), we get,
– 2 (m + n) + 2 m – m n = 0 or – 2 n – m n = 0
⇒ n (2 + m) = 0
Either n = 0
i.e., 0 l + 0 m + n = 0
Also,
l + m + n = 0
Solving, we get,
$fraction numerator straight l over denominator 0 minus 1 end fraction space equals space fraction numerator straight m over denominator 1 minus 0 end fraction space equals fraction numerator straight n over denominator 0 minus 0 end fraction$

$therefore$ $straight l over 1 space equals space fraction numerator straight m over denominator negative 1 end fraction space equals space straight n over 0$
or 2 + m = 0 i.e., m = -2
$therefore$ from (1), l - 2 + n = 0
Now, l = 1, n = 1 satisfy it
Also, l = 1, m =-2, n = 1 satisfy (2)
$therefore$ we have
l = 1, m = -2, n = 1
∴ direction-ratios of two lines are 1, – 1, 0 and 1, –2, 1.
Let θ be the angle between the lines
$therefore space space space cos space straight theta space equals space fraction numerator left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis negative 2 right parenthesis space plus space left parenthesis 0 right parenthesis thin space left parenthesis 1 right parenthesis over denominator square root of 1 plus 1 plus 0 end root space square root of 1 plus 4 plus 1 end root end fraction space equals space fraction numerator 1 plus 2 plus 0 over denominator square root of 2 square root of 6 end fraction space equals space fraction numerator 3 over denominator 2 square root of 3 end fraction$
$therefore space space cos space straight theta space equals space fraction numerator square root of 3 over denominator 2 end fraction space space space rightwards double arrow space space space space straight theta space equals space 30 space degree$

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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