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The vector equation of the line x-53=y+47=z-62 is ______.

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Question

The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.

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Solution

The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = 5hati - 4hatj + 6hatk + lambda(3hati + 7hatj + 2hatk)`.

Explanation:

We have, `(x - 5)/3 = (y + 4)/7 = (z - 6)/2`

The given line passes through the point (5, - 4 , 6 ) and has direction ratios proportional to 3, 7, 2.

Vector equation of the given line passing through the point having position vector

`veca = 5hati - 4hatj + 6hatk` and parallel to a vecto `vecb = 3hati + 7hatj + 2hatk` is `vecr = veca + lambdavecb`

⇒ `vecr = 5hati - 4hatj + 6hatk + lambda(3hati + 7hatj + 2hatk)`

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Chapter 12: Introduction to Three Dimensional Geometry - Exercise [Page 239]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 12 Introduction to Three Dimensional Geometry
Exercise | Q 39 | Page 239

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