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Tamil Nadu Board of Secondary EducationHSC Science Class 12

Find the parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (1, – 2, 3) and parallel to the straight line passing through the points

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Question

Find the parametric form of vector equation and Cartesian equations of the plane passing through the points (2, 2, 1), (1, – 2, 3) and parallel to the straight line passing through the points (2, 1, – 3) and (– 1, 5, – 8)

Sum
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Solution

`vec"a" = 2hat"i" + 2hat"j" + hat"k"`

`vec"b" = hat"i" - 2hat"j" + 3hat"k"`

`vec"c" = (2 + 1)hat"i" + (1 - 5)hat"j" + (-3 + 8)hat"k" = 3hat"i" - 4hat"j" + 5hat"k"`

`vec"r" = vec"a" + "s"(vec"b" - vec"a") + "t"vec"c"`

`vec"b" - vec"a" = -hat"i" - 4hat"j" + 2hat"k"`

Parametric form

`vec"r" = (2hat"i" + 2hat"j" + hat"k") + "s"(-hat"i" - 4hat"j" + 2hat"k") + "t"(3hat"i" - 4hat"j" + 5hat"k")`

`(vec"b" - vec"a") xx vec"c" = |(hat"i", hat"j", hat"k"),(-1, -4, 2),(3, -4, 5)|`

= `hat"i"(-20 + 8)  hat"j"(- 5 - 6) + hat"k"(4 + 12)`

= `-12hat"i" + 11hat"j" + 16hat"k"`

`[vec"r" - (2hat"i" + 2hat"j" + hat"k")]*(-12hat"i" + 11hat"j" + 16hat"k")` = 0

`vec"r"*(-12hat"i" + 11hat"j" + 16hat"k") = - 24 + 22 + 16` = 14

Cartesian equation

– 12x + 11y + 16z = 14

12x – 11y – 16z = – 14

12x – 11y – 16z + 14 = 0

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Different Forms of Equation of a Plane
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Chapter 6: Applications of Vector Algebra - Exercise 6.7 [Page 263]

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Samacheer Kalvi Mathematics - Volume 1 and 2 [English] Class 12 TN Board
Chapter 6 Applications of Vector Algebra
Exercise 6.7 | Q 3 | Page 263

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