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

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 Multiple Choice QuestionsShort Answer Type

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101.

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


Let A (1, 2, 4), B (– 2, 1, 2), C (2, 4, – 3) be vertices of Δ ABC.

AB space equals space square root of left parenthesis negative 2 minus 1 right parenthesis squared plus left parenthesis 1 minus 2 right parenthesis squared plus left parenthesis 2 minus 4 right parenthesis squared end root space equals space square root of 9 plus 1 plus 4 end root space equals space square root of 14
AC space equals space square root of left parenthesis 2 minus 1 right parenthesis squared plus left parenthesis 4 minus 2 right parenthesis squared plus left parenthesis negative 3 minus 4 right parenthesis squared end root space space equals square root of 1 plus 4 plus 49 end root space equals space square root of 54

Direction-ratios of AB are – 2, – 1, 1 – 2, 2 – 4 i.e., – 3, – 1,– 2 respectively.
Direction-ratios of AC are 2 – 1, 4 – 2, – 3 – 4 i.e., 1, 2, – 7 respectively.
cos space straight A space equals space fraction numerator left parenthesis negative 3 right parenthesis thin space left parenthesis 1 right parenthesis space plus space left parenthesis negative 1 right parenthesis thin space left parenthesis 2 right parenthesis space plus left parenthesis negative 2 right parenthesis thin space left parenthesis negative 7 right parenthesis over denominator square root of 9 plus 1 plus 4 end root square root of 1 plus 4 plus 49 end root end fraction space equals fraction numerator negative 3 minus 2 plus 14 over denominator square root of 14 space square root of 54 end fraction space equals space fraction numerator 9 over denominator square root of 14 space square root of 54 end fraction
sin space straight A space equals space square root of 1 minus cos squared straight A end root space equals space square root of 1 minus fraction numerator 81 over denominator 14 cross times 54 end fraction end root space equals space square root of fraction numerator 766 minus 81 over denominator 14 cross times 54 end fraction end root space equals fraction numerator square root of 675 over denominator square root of 14 space square root of 54 end fraction
Area space of increment thin space ABC space equals space 1 half space AB. space AC. space sin space straight A space equals space 1 half square root of 14. space square root of 54. space fraction numerator square root of 675 over denominator square root of 14 space square root of 54 end fraction
                          equals space 1 half square root of 675 space sq. space units. space




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 Multiple Choice QuestionsLong 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–1open 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
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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 straight u squared over straight a plus straight v squared over straight b plus straight w squared over straight c equals 0.

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