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If P, Q and R are three points having coordinates (3, –2, –1), (1, 3, 4), (2, 1, –2) in XYZ space, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by  
  • a)
    3      
  • b)
    5  
  • c)
    7      
  • d)
    9
Correct answer is option 'A'. Can you explain this answer?
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If P, Q and R are three points having coordinates (3, –2, –...
The equation of the plane OQR is (O being origin). 
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If P, Q and R are three points having coordinates (3, –2, –...
Calculating the distance from point to plane in XYZ space

Introduction: In XYZ space, given three points P, Q, and R, we need to calculate the distance from point P to plane OQR (O being the origin of the coordinate system).

Solution:
1. Find the normal vector to the plane OQR using cross product of vectors OQ and OR.
2. Normalize the normal vector to get a unit vector.
3. Use dot product of the unit normal vector and vector OP to calculate the distance from point P to plane OQR.

Given points P, Q, and R:
P(3, 2, 1), Q(1, 3, 4), R(2, 1, 2)

1. Find the normal vector to the plane OQR:
OQ = Q - O = (1, 3, 4)
OR = R - O = (2, 1, 2)
N = OQ x OR = (-10, 6, -5)

2. Normalize the normal vector:
|N| = sqrt((-10)^2 + 6^2 + (-5)^2) = sqrt(181)
n = N/|N| = (-10/sqrt(181), 6/sqrt(181), -5/sqrt(181))

3. Calculate the distance from point P to plane OQR:
OP = P - O = (3, 2, 1)
d = |OP . n| = |(3, 2, 1) . (-10/sqrt(181), 6/sqrt(181), -5/sqrt(181))| = 3

Therefore, the distance from point P to plane OQR is 3.

Conclusion: The correct option is A (3).
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If P, Q and R are three points having coordinates (3, –2, –1), (1, 3, 4), (2, 1, –2) in XYZ space, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by a)3 b)5 c)7 d)9Correct answer is option 'A'. Can you explain this answer?
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