Unit Vector Perpendicular to a Triangle?

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To find a unit vector perpendicular to triangle PQS, the discussion emphasizes the use of the cross product of two vectors formed by the triangle's vertices. The participants suggest calculating the vector difference between points P and S, resulting in a vector of <-1, -3, -7>, which can then be crossed with vector Q. It is noted that the cross product yields a vector perpendicular to both input vectors, making it suitable for finding the desired perpendicular vector. Clarification is provided that the dot product results in a scalar, not a vector, reinforcing the focus on the cross product for this problem. The conversation concludes with a consideration of whether vector Q is parallel to the triangle.
neotriz
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Homework Statement



Given: P=(1,-2,3), Q=(-4,2,5) and S=(2,1-4)

Find a unit vector that is perpendicular to triangle PQS


Homework Equations



Cross and Dot Product

The Attempt at a Solution




Correct me if I'm doing wrong. I have two solutions that I've thought:

1)What I would do is find the dot product of S and P and using that result, cross product with Q.

or

2)Find two dot products of two opposite triangle sidesand cross product on those.


Just want to make sure I am doing right
 
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If you take a dot product, you're left with a scalar. The cross product requires vectors to be calculated.

You have the right idea of using the cross product though. One of the fundamental properties of the cross product is that v_1 \times v_2 is perpendicular to both v1 and v2. So you want to find two vectors from your triangle such that, if you get a vector perpendicular to both of them, you get a vector perpendicular to the triangle.
 
I forgot that in dot product it gives you scalar result

How about this then:

I find P to S vector difference , which will result a vector of <-1, -3, -7> and using that vector, I cross product with Q
 
neotriz said:
I forgot that in dot product it gives you scalar result

How about this then:

I find P to S vector difference , which will result a vector of <-1, -3, -7>

This is good. Taking the difference between S and P gives a vector that's pointing along one of the edges of the triangle, so is parallel to the triangle

and using that vector, I cross product with Q

On the other hand, is Q parallel to the triangle?
 
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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