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Introduction To Three Dimensional Geometry

Short Answer Type

71. Find the co-ordinates of the point which divides the line segment formed by joining the points (- 2, 3, 5) and (1, -4, 6) in the ratio of 2 : 3 externally.

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72. Given that P (3, 2, -4), Q (5, 4, -6) and R (9, 8, -10) are collinear. Find the ratio in which Q divides PR.

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73. Using section formula, prove that the three points (-4, 6, 10), (2, 4, 6) and (14, 0, - 2) are collinear. Also find the ratio in which point B divides the join of A and C.

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74. Find the co-ordinates of the points of trisection of the line segment joining the points P (3, 2, - 1) and Q (1, 2, 5).

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75.

Find the ratio in which the YZ-plane divides the line segment formed by joining the points (–2, 4, 7) and (3, –5, 8).

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76.

Find the ratio in which the join of P (2, 1, -1) and Q (3, 2, 4) is divided by XY-plane.

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77. A point R with x-coordinate 4 lies on the line joining the points P (2, - 3, 4) and Q (8, 0, 10). Find the co-ordinates of point R.

Let point R (4, y, z) divide the join of P (2, - 3, 4) and Q (8, 0, 10) in the ratio k:1.
$rightwards double arrow$ R is $open parentheses fraction numerator 8 straight k plus 2 over denominator straight k plus 1 end fraction comma space fraction numerator 0 straight k minus 3 over denominator straight k plus 1 end fraction comma space fraction numerator 10 straight k plus 4 over denominator straight k plus 1 end fraction close parentheses$
∴ $open parentheses fraction numerator 8 straight k plus 2 over denominator straight k plus 1 end fraction comma space fraction numerator negative 3 over denominator straight k plus 1 end fraction comma space fraction numerator 10 straight k plus 4 over denominator straight k plus 1 end fraction close parentheses space left right arrow space left parenthesis 4 comma space straight y comma space straight z right parenthesis$

$rightwards double arrow$ $fraction numerator 8 straight k plus 2 over denominator straight k plus 1 end fraction equals 4 space rightwards double arrow space 8 straight k plus 2 space equals space 4 straight k plus 4 space rightwards double arrow space 4 straight k space equals space 2 space rightwards double arrow space straight k equals 1 half$
∴ $straight y equals fraction numerator negative 3 over denominator begin display style 1 half end style plus 1 end fraction equals space minus 3 cross times 2 over 3 equals negative 2 comma space straight z space equals space fraction numerator 10 straight k plus 4 over denominator straight k plus 1 end fraction equals fraction numerator 10 open parentheses begin display style 1 half end style close parentheses plus 4 over denominator begin display style 1 half end style plus 1 end fraction equals fraction numerator 9 over denominator begin display style 3 over 2 end style end fraction equals 6$
Hence, point R is (4, -2, 6).

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Long Answer Type

78. Find the lengths of the medians of the triangle with vertices A (0, 0, 6), B (0, 4, 0) and C (6, 0, 0).

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Short Answer Type

79. A (3, 2, 0), B (5, 3, 2) and C (-9, 6, -3) are the vertices of a triangle ABC. The bisector AD of ∆ BAC meets BC at D. Find the co-ordinates of D.

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80. The co-ordinates of the mid-point C of a line segment AB are (-1, -2, 1). If the co-ordinates of point A are (2, 1, -3), find the co-ordinates of B.

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