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Three Dimensional Geometry

Question
CBSEENMA12033311

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

Solution
Let the given line AB be
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

Any point M on this line is (2 r + 1, – 3 r – 1, 8 r – 10)
Let this point M be the foot of perpendicular form P( 1,0, 0) on AB.
Direction ratios of PM are
2 r + 1 – 1, –3 r – 1 – 0,
8 r – 10 – 0 i.e. 2 r, – 3 r – 1, 8 r – 10
Direction-ratios of AB are 2, –3, 8
Since PM ⊥ AB
∴  (2 r) (2) + (– 3r – 1) (–3) + (8 r – 10) (8) = 0
∴  4 r + 9 r + 3 + 64 r – 80 = 0
∴ 77 r = 77 ⇒ r = 1
∴ M is (3, –4, –2)
Required distance  = PM = square root of left parenthesis 3 minus 1 right parenthesis squared plus left parenthesis 4 minus 0 right parenthesis squared plus left parenthesis negative 2 minus 0 right parenthesis squared end root
                               equals space square root of 4 plus 16 plus 4 end root space equals space square root of 24 space equals space square root of 4 cross times 6 end root space equals space 2 square root of 6.

 

Some More Questions From Three Dimensional Geometry Chapter

Find the direction cosines of x, y and z-axis.