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in Three Dimensional Geometry by (28.9k points)
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Consider three points (6, -1, 1), (5, 1, 2) and (1, – 5, -4) on space.

  1. Find the Cartesian equation of the plane passing through these points. 
  2. Find direction ratios normal to the Plane.
  3. Find a unit vector normal to the Plane. 

1 Answer

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Best answer

1. Equation of a plane passing through the points (6, -1, 1),(5, 1, 2) and (1, -5, 4)

⇒ (x – 6)(-10 + 4) – (y + 1)(5 + 5) + (z – 1)(4 + 10) = 0

⇒ (x – 6)(-6) – (y + 1)(10) + (z – 1)(14) = 0

⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0

⇒ 6x +10y – 14z -12 = 0.

2. Dr’s normal to the plane are 6 : 10 : -14 ⇒ 3 : 5 : -7.

3. Since the dr’s normal to the plane are 3 : 5 : -7, a unit vector in this direction is

\(\frac{3i+5j-7k}{\sqrt{3^2+5^2+(-7)^2}} = \frac{3i+5j-7k}{\sqrt{83}}\)

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