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brinlin
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MHB Components, Projection, and Resolution
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The discussion focuses on understanding vector components, specifically how to determine the component of one vector along another using trigonometry. It explains that by drawing a perpendicular from the tip of vector u to vector v, one can form a right triangle where the hypotenuse is the length of u. The length of the component of u along v corresponds to the adjacent side of this triangle, which can be calculated using the cosine of the angle between the two vectors. The dot product formula, u·v = |u||v| cos(θ), is highlighted as a useful tool for these calculations. Understanding these concepts is essential for solving vector-related problems effectively.
- #2
HOI
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Where are you getting these problems? You seem to be saying you know nothing at all about trigonometry!
Imagine the vectors u and v with their "tails" together. Draw a perpendicular from the tip of u to v. The "component of u along v" is the distance from the common tails to that perpendicular. So you have a right triangle where the length of the hypotenuse is the length of u and one angle is the angle between u and v. The "component of u along v" is the length of the "near side" of that triangle so you will need the cosine of the two vectors. If you know that the dot product of two vectors is $u\cdot v= |u||v| cos(\theta)$ then that will be easy to calculate.
Imagine the vectors u and v with their "tails" together. Draw a perpendicular from the tip of u to v. The "component of u along v" is the distance from the common tails to that perpendicular. So you have a right triangle where the length of the hypotenuse is the length of u and one angle is the angle between u and v. The "component of u along v" is the length of the "near side" of that triangle so you will need the cosine of the two vectors. If you know that the dot product of two vectors is $u\cdot v= |u||v| cos(\theta)$ then that will be easy to calculate.
For the sake of argument, lets assume this is, in fact, a right triangle with a rectangle inscribed in it.
Here is what looks like a very elegant solution:
Triangles A and B are similar
a/4=2/b
ab=8
But what is happening in that second line? Is there a property of similar triangles I'm forgetting?
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