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

MATH152 April 2010

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Question B 06 (a)
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 area of T?

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?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Don't fall back to the elementary method of calculating base and height. There is a linear-algebraic formula for area of parallelogram which you can make use of.

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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. The area of T is half of the area of a parallelogram with vertices (0, 1, 2), (1, 1, 5) and (-1, 2, 2). Meanwhile, the area of a parallelogram is the magnitude of the cross product of vectors of two adjacent sides. A convenient choice of vectors representing two adjacent sides are the ones with common vertex (0, 1, 2). They are (1, 1, 5) - (0, 1, 2) = (1, 0, 3) and (-1, 2, 2) - (0, 1, 2) = (-1, 1, 0) Hence, $T={\frac {1}{2}}\left|(1,0,3)\times (-1,1,0)\right|=1/2\left|(-3,-3,1)\right|={\frac {\sqrt {19}}{2}}$