Science:Math Exam Resources/Courses/MATH152/April 2012/Question 01 (d)

MATH152 April 2012

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Question 01 (d)
Let $S_{1}$ be the plane perpendicular to $\mathbf {n} =[2,1,-1]$ and through the point P=[1,-1,0]. Let $S_{2}$ be the plane through the points A=[1,0,0], B=[-1,2,1], C=[0,1,1]. By L, we denote the line of intersection of $S_{1}$ and $S_{2}$. Calculate the area of the triangle ABC.

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
How can we relate the area of a triangle formed between two vectors to the cross product?

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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 the parallelogram spanned by two vectors is the magnitude of their cross product. The area of a triangle is half the area of a parallelogram. The parallelogram form by A, B, and C is spanned by ${\vec {AB}}$ and ${\vec {AC}}$ which in 1(a), we already determined were ${\begin{aligned}{\vec {AB}}&=[-2,2,1]\\{\vec {AC}}&=[-1,1,1].\end{aligned}}$ We need the cross product of this, but as we saw in 1(a), this is just the normal vector to the plane $S_{2}$ and therefore ${\vec {AB}}\times {\vec {AC}}=\mathbf {m} =[1,1,0]$. Therefore the area of the triangle is ${\textrm {area}}={\frac {1}{2}}||{\vec {AB}}\times {\vec {AC}}||={\frac {1}{2}}||\mathbf {m} ||={\frac {1}{2}}{\sqrt {1^{2}+1^{2}+0^{2}}}={\frac {1}{2}}{\sqrt {2}}.$