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What will be the value of 'P' so that the lines 1-x3=7y-142P=z-32 and 7-7x3P=y-51=6-z5 at right angles. -

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प्रश्न

What will be the value of 'P' so that the lines `(1 - x)/3 = (7y - 14)/(2P) = (z - 3)/2` and `(7 - 7x)/(3P) = (y - 5)/1 = (6 - z)/5` at right angles.

विकल्प

  • 11

  • 70

  • `70/11`

  • `11/70`

MCQ

उत्तर

`70/11`

Explanation:

The given equations are not in standard form, but they can be written as follows:

`(x - 1)/(-3) = (y - 2)/((2P)/7) = (z - 3)/2`  .......(i)

`(x - 1)/((3P)/7) = (y - 5)/1 = (z - 6)/(-5)`  .......(ii)

The direction ratios of the given lines are `-3, (2P)/7, 2` and  `(-3P)/7, 1, -5`.

The lines are perpendicular to one another after that.

`(-3) ((-3P)/7) + ((2P)/7) (1) + (2) (-5)` = 0

9P + 2P – 70 = 0

11P = 70

⇒ P = `70/11`

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