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Three Dimensional Geometry

Question
CBSEENMA12033316

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

Solution
The equation of line BC
or   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 space equals space straight k

Any point D on it is
(4 k + 1, 4, – 2 k + 6)
Let D be foot of perpendicular from A on BC.
Direction ratios of AD are
4 k + 1–1, 4 – 2, – 2 k + 6 – 1 i.e., 4 k , 2, – 2 k + 5
Direction ratios of BC are 4, 0, – 2
Since AD is perpendicular to BC
∴    (4 k) (4) + (2) (0) + (– 2 k + 5) (–2) = 0
therefore space space space space space 16 straight k plus 0 plus 4 straight k minus 10 space equals space 0 space space space space space space space rightwards double arrow space space space 20 space straight k space equals space space 10 space space space space space space space rightwards double arrow space space space straight k space equals space 1 half
therefore space space space space space straight D space is space left parenthesis 2 plus 1 comma space 4 comma space minus 1 comma space plus 6 right parenthesis space straight i. straight e. comma space left parenthesis 3 comma space 4 comma space 5 right parenthesis