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in Three Dimensional Geometry by (28.2k points)
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Consider three points on space ( 2, 1, 0 ), (3, -2, -2) and (3, 1, 7)

  1. Find the Cartesian equation of the plane passing through the above points. 
  2. Convert the above equation into vector form.
  3. Hence, find a unit vector perpendicular to the above plane and also find the perpendicular distance of the plane from the origin

1 Answer

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Best answer

1. Equation of the plane is

⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0

⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0

⇒ 7x + 3y – z = 17.

2. Vector form is \(\bar r\).(7i + 3j – k) = 17 _____(1)

3. Now, |7i + 3j – k| = \(\sqrt{49+9+1} = \sqrt{59}\)

Dividing equation (1) by \(\sqrt{59}\), we get

Therefore the above equation is the normal form of the plane. Then \(\frac{7i+3j-k}{\sqrt{59}}\)is the unit vector perpendicular to the plane and \(\frac{17}{\sqrt{59}}\) is the perpendicular distance from the origin.

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