Find the image of the point (1, 6, 3) in the line  from Mathem

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 Multiple Choice QuestionsShort Answer Type

91. Find the length of perpendicular from (2, 1, 3) on the line
fraction numerator straight x minus 4 over denominator 5 end fraction space equals fraction numerator straight y minus 2 over denominator 4 end fraction space equals space fraction numerator straight z minus 3 over denominator 3 end fraction.
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92.

Find the length and foot of the perpendicular drawn from the point (3, 4, 5) on the line 
fraction numerator straight x minus 2 over denominator 2 end fraction space equals space fraction numerator straight y minus 3 over denominator 5 end fraction space equals space fraction numerator straight z minus 1 over denominator 3 end fraction

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93.

Find the perpendicular distance of the point (1, 0, 0) form the line fraction numerator straight x minus 1 over denominator 2 end fraction space equals space fraction numerator straight y plus 1 over denominator negative 3 end fraction space equals space fraction numerator straight z plus 10 over denominator 8 end fraction

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94.

Find the length of the perpendicular drawn from the point (1, ,2 3) on the line
fraction numerator straight x minus 6 over denominator 3 end fraction space equals space fraction numerator straight y minus 7 over denominator 2 end fraction space equals space fraction numerator straight z minus 7 over denominator negative 2 end fraction

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95. Find the length and the foot of the perpendicular drawn from the point (2, – 1, 5) to the line fraction numerator straight x minus 11 over denominator 10 end fraction space equals space fraction numerator straight y plus 2 over denominator negative 4 end fraction space equals space fraction numerator straight z plus 8 over denominator negative 11 end fraction.
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 Multiple Choice QuestionsLong Answer Type

96. Find the equation of the perpendicular drawn from the point (2, 4, – 1) to the line  fraction numerator straight x plus 5 over denominator 1 end fraction space equals space fraction numerator straight y plus 3 over denominator 4 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 9 end fraction.
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97. Find the length of perpendicular from (3, 2, 1) on the line 
fraction numerator straight x minus 4 over denominator 5 end fraction space equals space fraction numerator straight y minus 2 over denominator 3 end fraction space equals space fraction numerator straight z minus 3 over denominator 4 end fraction

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 Multiple Choice QuestionsShort Answer Type

98. Find the coordinates of the foot of perpendicular drawn from the point A (1, 2, 1) to the line
fraction numerator straight x minus 1 over denominator 4 end fraction space equals space fraction numerator straight y minus 4 over denominator 0 end fraction space equals space fraction numerator straight z minus 6 over denominator negative 2 end fraction
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 Multiple Choice QuestionsLong Answer Type

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99. Find the image of the point (1, 6, 3) in the line straight x over 1 space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 3 end fraction.


The equations of given line are
straight x over 1 space equals space fraction numerator straight y minus 1 over denominator 2 end fraction space equals space fraction numerator straight z minus 2 over denominator 3 end fraction

From P (1, 6, 3) , draw PM ⊥ AB and produce it to P' (α, β γ) such that M is mid-point of PP'. Then P' is image of P in line AB.
Any point M as line AB is
(r. 2 r + 1, 3 r + 2)
Direction ratios of AB are 1, 2, 3
Direction ratios of PM are r – 1 , 2 r + 1 – 6 , 3 r + 2 – 3
i.e. r – 1, 2 r – 5, 3 r – 1
∵ PM ⊥ AB∴ (1) (r – 1) + (2) (2 r – 5) + (3) (3 r – 1) = 0
∴ i + 4 r – 10 + 9 r – 3 = 0
∴ 14 r = 14 ⇒ r = 1
∴ M is (1, 3, 5)
Now M is mid-point of PP'
therefore space space space space space fraction numerator straight alpha plus 1 over denominator 2 end fraction space equals space 1 comma space space space space fraction numerator straight beta plus 6 over denominator 2 end fraction space equals space 3 comma space space space fraction numerator straight gamma plus 3 over denominator 2 end fraction space equals space 5
therefore space space space straight alpha plus 1 space equals space 2 comma space space space space space space space space space straight beta plus 6 space equals space 6 comma space space space space straight gamma space plus space 3 space equals space 10
therefore space space space straight alpha space equals space 1 comma space space space space space space space space space space space straight beta space equals space 0 comma space space space space space space space space straight gamma space equals space 7
therefore space space space space image space is space left parenthesis 1 comma space 0 comma space 7 right parenthesis

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 Multiple Choice QuestionsShort Answer Type

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