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Find the Vector and the Cartesian Equations of the Lines that Pass Through the Origin and (5, −2, 3).

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Question

Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).

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Solution

The required line passes through the origin. Therefore, its position vector is given by,

`veca = vec0`  ...(1)

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation, `vecb = 5hati - 2hatj + 3hatk`

The equation of the line in vector form through a point with position vector `veca` and parallel to `vecb` is, `vecr = veca + lambdavecb`,  `lambda in R`

The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given by `(x-x_1)/x = (y-y_1)/b  = (z-z_1)/c`

Therefore, the equation of the required line in the Cartesian form is

 

 

 

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Chapter 11: Three Dimensional Geometry - Exercise 11.1 [Page 477]

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NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 11 Three Dimensional Geometry
Exercise 11.1 | Q 8 | Page 477

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