2007 AMC 12B Problems/Problem 25
Contents
Problem
Points and are located in 3-dimensional space with and . The plane of is parallel to . What is the area of ?
Solution
Let , and . Since , we could let , , and . Now to get back to we need another vertex . Now if we look at this configuration as if it was two dimensions, we would see a square missing a side if we don't draw . Now we can bend these three sides into an equilateral triangle, and the coordinates change: , , , , and . Checking for all the requirements, they are all satisfied. Now we find the area of triangle . The side lengths of this triangle are , which is an isosceles right triangle. Thus the area of it is .
Solution 2
Similar to solution 1, we allow , , and . This creates the isosceles right triangle on the plane of
Now, note that . This means that there exists some vector parallel to the plane of that forms two right angles with and . By definition, this is the cross product of the two vectors and . Finding this cross product, we take the determinant of vectors
and
*Note that z is constant because the line is parallel to the plane*
to get
Because there can be no movement in the direction, the k unit vector must be zero. Also, because the i unit vector must be orthogonal and also 0. Thus, the vector of line is simply
From this, you can figure out that line , and the area of .
See also
2007 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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All AMC 12 Problems and Solutions |
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