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Find the distance of the point (– 2, 4, – 5) from the line x+33=y-45=z+86 - Mathematics

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Question

Find the distance of the point (– 2, 4, – 5) from the line `(x + 3)/3 = (y - 4)/5 = (z + 8)/6`

Sum

Solution

Here P (–2, 4, – 5) is the given point.

Any point Q on the line is given by `(3lambda - 3, 5lambda + 4, (6lambda - 8)`

`vec"PQ" = (3lambda - 1) hati + 5lambdahatj + (6lambda - 3)hatk`.

Since `vec"PQ" ⊥ (3hati + 5hatj + 6hatk)`, we have

`3(3lambda - 1) + 5(5lambda) + 6(6lambda - 3)` = 0

`9lambda + 25lambda + 36lambda` = 21

i.e. `lambda = 3/10`

Thus `vec"PQ" = - 1/10 hati + 15/10 hatj - 12/10 hatk`

Hence `|vec"PQ"| = 1/10 sqrt(1 + 225 + 144)`

= `sqrt(37/10)`.

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Chapter 12: Introduction to Three Dimensional Geometry - Solved Examples [Page 226]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 12 Introduction to Three Dimensional Geometry
Solved Examples | Q 6 | Page 226

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