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Three Dimensional Geometry

Question
CBSEENMA12033306

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.

Solution
The equation of line through B (1, 4, 6) and C (5, 4, 4) is
                      fraction numerator straight x minus 1 over denominator 5 minus 1 end fraction space equals fraction numerator straight y minus 4 over denominator 4 minus 4 end fraction space equals space fraction numerator straight z minus 6 over denominator 4 minus 6 end fraction
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
therefore space space space space space space left parenthesis 4 space straight k right parenthesis thin space left parenthesis 4 right parenthesis space plus space left parenthesis 2 right parenthesis thin space left parenthesis 0 right parenthesis space plus space left parenthesis negative 2 space straight k space plus 5 right parenthesis thin space left parenthesis negative 2 right parenthesis space equals space 0
therefore space space space space space 16 space straight k space plus space 0 space plus space 4 space straight k space minus space 10 space equals space 0 space space space space space space rightwards double arrow space space space 20 space straight k space equals space 10 space space space space space space space rightwards double arrow space space straight k space equals space 1 half
therefore space space space space straight D space is space space left parenthesis 2 plus 1 comma space space 4 comma space space minus 1 comma space plus 6 right parenthesis space straight i. straight e. space left parenthesis 3 comma space 4 comma space 5 right parenthesis
⊥ distance of A from BC = distance AD
equals space square root of left parenthesis 3 minus 1 right parenthesis squared plus left parenthesis 4 minus 2 right parenthesis squared plus left parenthesis 5 minus 1 right parenthesis squared end root space equals space square root of 4 plus 4 plus 16 end root space equals square root of 24 space equals space 2 square root of 6