Finding an Orthogonal Vector and Calculating Triangle Area from Given Points

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To find a nonzero vector orthogonal to the plane formed by points P, Q, and R, the cross product of vectors PQ and RP is essential. The vectors are calculated as PQ = <4, 3, -2> and RP = <5, 5, 1>. Computing the cross product will yield a vector perpendicular to the plane. The area of triangle PQR can then be determined using the magnitude of this cross product. Understanding the geometric definition of the cross product is crucial for solving the problem effectively.
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Homework Statement

Find a nonzero vector orthogonal to the plane through points P (0, -2, 0) Q (4, 1, -2) and R (5,3,1) and find the area of the triangle formed by PQR.

The attempt at a solution
To be honest, I am not entirely sure how to do this problem. I've looked through my textbook and notes, but there is no example that is of the same form of this problem. However, I suspect the cross product is important:

PQ has a vector of <4,3,-2>
RP has a vector of <5,5,1>

Trying to find the cross product I get:
(3--6) - (4-10) + (20 - 15)
Equals 20.

Is that right, and what do I do from here?
 
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You're on the right track. Those 2 vectors you have are parallel to your plane. If you compute the cross product between them, you will get a new vector perpendicular to the 2 vectors and hence perpendicular to the plane. Do you know how to compute the cross product? What you get should be a vector. The next bit relies on the geometric definition of the cross product.

Go over on computing the cross product as it's all you'll need.
http://en.wikipedia.org/wiki/Cross_product
 
Wow, epic fail. Thanks!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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