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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

The given equation are
2 l – m + 2 n = 0    ...(1)
and m n + n l + l m = 0    ....(2)
From (1), m = 2 l + 2 n    ....(3)
From (2) and (3), we get,
n (2 l + 2 n) + n l + l (2 l + 2 n) = 0
or 2 n l + 2 n² + n l + 2 l² + 2 n l = 0 or 2 l² + 5 l n + 2 n² = 0
⇒ (2 l + n) (l + 2 n) = 0
$therefore$ either 2l+ n = 0
i.e. 2l + 0 m + n = 0
Also, 2l - m + 2n = 0
Solving, we get,
$fraction numerator straight l over denominator 0 plus 1 end fraction space equals space fraction numerator straight m over denominator 2 minus 4 end fraction space equals space fraction numerator straight n over denominator negative 2 minus 0 end fraction$

$therefore space space space space space straight l over straight l space equals space fraction numerator straight m over denominator negative 2 end fraction space equals space fraction numerator straight n over denominator negative 2 end fraction$

or l + 2n = 0
i.e., l + 0 m + 2 n = 0
Also,
2l - m + 2n = 0
Solving, we get,
   $fraction numerator straight l over denominator 0 plus 2 end fraction space equals space fraction numerator straight m over denominator 4 minus 2 end fraction space equals space fraction numerator straight n over denominator negative 1 minus 0 end fraction$

$therefore$ $straight l over 2 space equals space straight m over 2 space equals space fraction numerator straight n over denominator negative 1 end fraction$

∴ direction-ratios of two lines are 1, – 2, – 2 and 2, 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 2 right parenthesis space plus space left parenthesis negative 2 right parenthesis thin space left parenthesis 2 right parenthesis space plus space left parenthesis negative 2 right parenthesis thin space left parenthesis negative 1 right parenthesis over denominator square root of 1 plus 4 plus 4 end root space square root of 4 plus 4 plus 1 end root end fraction fraction numerator 2 minus 4 plus 2 over denominator 3 cross times 3 end fraction space equals space 0

∴ θ = 90°

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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