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Three Dimensional Geometry

Short Answer Type

81. If the lines

fraction numerator straight x minus 1 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 2 over denominator 2 straight k end fraction space equals space fraction numerator straight z minus 3 over denominator 2 end fraction space and space fraction numerator straight x minus 1 over denominator 3 straight k end fraction space equals space fraction numerator straight y minus 1 over denominator 1 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 5 end fraction

are prependicular, then find the value of k.

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82.

Show that the lines x = ay + b, z = cy + d and x = a' y + b' , z = c' y + d' are perpendicular to each other, if aa' + cc' + 1 = 0.

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83. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, –6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

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Long Answer Type

84. Find the equations of the line passing through the point P(–1, 3, –2) and perpendicular to the lines

straight x over 1 space equals space straight y over straight z space equals space straight z over 3 space space and space fraction numerator straight x plus 2 over denominator negative 3 end fraction space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z plus 1 over denominator 5 end fraction

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85. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

fraction numerator straight x minus 8 over denominator 3 end fraction space space equals space fraction numerator straight y plus 19 over denominator negative 16 end fraction space equals space fraction numerator straight z minus 10 over denominator 7 end fraction space and space fraction numerator straight x minus 15 over denominator 3 end fraction space equals space fraction numerator straight y minus 29 over denominator 8 end fraction space equals space fraction numerator straight z minus 5 over denominator negative 5 end fraction.

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86. Show that, if the axes are rectangular, then the equations of the line through (α, β, γ) right angles to the lines

straight x over straight l subscript 1 space equals space straight y over straight m subscript 1 space equals space straight z over straight n subscript 1 comma space space straight x over straight l subscript 2 space equals space straight y over straight m subscript 2 space equals space straight z over straight n subscript 2

are

fraction numerator straight x minus straight alpha over denominator straight m subscript 1 space straight n subscript 2 minus straight m subscript 2 straight n subscript 1 end fraction space equals space fraction numerator straight y minus straight beta over denominator straight n subscript 1 straight l subscript 2 minus straight n subscript 2 straight l subscript 1 end fraction space equals fraction numerator straight z minus straight gamma over denominator straight l subscript 1 straight m subscript 2 minus straight l subscript 2 straight m subscript 1 end fraction.

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87. If the straight lines having direction cosines given by al + bm + cn = 0 and fmn + gnI + hIm = 0 are perpendicular, then show that $straight f over straight a plus straight g over straight b plus straight h over straight c space equals space 0$

Given that a I + b m + c n = 0,
i.e.

straight n space equals fraction numerator negative left parenthesis al plus bm right parenthesis over denominator straight c end fraction

...(1)
Also,

straight f space straight m space straight n plus straight g space straight n space straight l plus straight h space straight l space straight m space equals space 0

...(2)
Substituting value of n from (1) in (2), we get

fm open square brackets fraction numerator negative left parenthesis al plus bm right parenthesis over denominator straight c end fraction close square brackets space plus space straight g space straight l space open square brackets fraction numerator negative left parenthesis al space plus space bm right parenthesis over denominator straight c end fraction close square brackets space plus space straight h space straight l space straight m space equals space 0

straight a space straight f space straight m space straight l space plus straight b space straight f space straight m squared space plus space straight a space straight g space straight l squared space plus space straight b space straight g space straight l space straight m space minus space straight h space straight c space straight l space straight m space equals space 0

On dividing both sides by m², we have

straight a space straight g space open parentheses straight l over straight m close parentheses squared plus straight l over straight m left parenthesis straight a space straight f space plus space straight b space straight g space minus space straight c space straight h right parenthesis space plus space straight b space straight f space equals space 0

If l₁, m₁, n₁ and l₂, m₂ , n₂are the direction cosines of the two lines, then the roots of the equation (3) as

straight l subscript 1 over straight m subscript 1 space and space straight l subscript 2 over straight m subscript 2 space space give

fraction numerator straight l subscript 1 space straight l subscript 2 over denominator straight m subscript 1 space straight m subscript 2 end fraction space equals space fraction numerator straight b space straight f over denominator straight a space straight g end fraction space space space straight i. straight e. comma space space space space fraction numerator straight l subscript 1 space straight l subscript 2 over denominator begin display style straight f over straight a end style end fraction space equals space fraction numerator straight m subscript 1 space straight m subscript 2 over denominator begin display style straight g over straight b end style end fraction

...(4)
Similarly, using (1) and (2) and by elimination of 1, we get

fraction numerator straight m subscript 1 space straight m subscript 2 over denominator begin display style straight g over straight b end style end fraction space equals space fraction numerator straight n subscript 1 space straight n subscript 2 over denominator begin display style straight h over straight c end style end fraction

...(5)
Combining (4) and (5), we have

fraction numerator straight l subscript 1 straight l subscript 2 over denominator begin display style straight f over straight a end style end fraction space equals space fraction numerator straight m subscript 1 space straight m subscript 2 over denominator begin display style straight g over straight b end style end fraction space equals space fraction numerator straight n subscript 1 space straight n subscript 2 over denominator begin display style straight h over straight c end style end fraction space equals space straight k space left parenthesis say right parenthesis

therefore space space space space straight l subscript 1 space straight l subscript 2 space equals space fraction numerator straight k space straight f over denominator straight a end fraction comma space space space space space straight m subscript 1 space straight m subscript 2 space equals space straight k space fraction numerator straight k space straight g over denominator straight b end fraction comma space space straight n subscript 1 straight n subscript 2 space equals space fraction numerator straight k space straight h over denominator straight c end fraction

...(6)
We know that the lines with direction cosines l₁, m₁, n₁ and l₂, m₂, n₂ are perpendicular if
l₁ l₂ + m₁m₂ + n₁n₂= 0 ....(7)
From (6) and (7), we get,

straight k open parentheses straight f over straight a close parentheses plus straight k open parentheses straight g over straight b close parentheses space plus space straight k space open parentheses straight h over straight c close parentheses space equals space 0 space space space space or space space space space straight f over straight a plus straight g over straight b plus straight h over straight c space equals space 0

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Short Answer Type

88. Find the coordinates of the foot of perpendicular drawn from the point A (1, 2, 1) to the line joining B (1, 4, 6) and (5, 4, 4). Also find the perpendicular distance of A from line BC.

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89. Find the co-ordinates of the foot of the perpendicular drawn from the point A(1, 8, 4) to the line joining the points B (0, –1, 3) and C (2, – 3, – 1).

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Long Answer Type

90. Find the foot of perpendicular from the point (0, 2, 3) on the line

fraction numerator straight x plus 3 over denominator 5 end fraction space equals fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z plus 4 over denominator 3 end fraction

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