Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 06 (b)

MATH152 April 2010

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Question B 06 (b)
Consider the triangle T in three dimensions with vertices (0, 1, 2), (1, 1, 5) and (−1, 2, 2). Consider also the plane P that contains T. What is the normal (perpendicular) direction to P?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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Hint
How does the cross product relate to perpendicular vectors?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. First pick any two linearly independent vectors lying in P (meaning the heads and tails of the vectors can be drawn in the plane P). A convenient choice will be two sides of the triangle T: (1, 1, 5) - (0, 1, 2) = (1, 0, 3) and (-1, 2, 2) - (0, 1, 2) = (-1, 1, 0). A normal vector to the plane is given by the cross product of these two vectors (which is calculated in part (a) already): ${\vec {N}}=(1,0,3)\times (-1,1,0)=(-3,-3,1)$ The above vector gives the normal direction.