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Topic: -=[EXAM PRACTISE]=- (Read 955 times) |
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Rego
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hi, im studying for my 2nd year degreee graphics course and i came across 2 practise exam questions which i am not able to solve! could you help?
1) Give a formula in terms of dot-product for the shortest distance from the origin to the line y=mx+c
2) Use the notion of a dot-product to explain why the inverse of a rotation matrix is the same as its transpose
thanks!!!
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Pietro K.C.
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Re: -=[EXAM PRACTISE]=-
« Reply #1 on: Apr 17th, 2004, 10:12am » |
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Hi Rego!
I'll try to give you some hints to solve #1. Call 'r' the line given by the equation 'y=mx+c', 'O' the origin, and try the following:
a. There is a shortest distance from O to r; that means there is a point on r which is closer to O than any other.
b. Draw a diagram of a generic line on the cartesian plane and pick what seems to be the point on it closest to O. Draw a line 's' from O to that point.
c. The angle between r and s - what kind of angle does that look like?
d. Prove the general case of your conjecture, i.e. that the lines r and s, so constructed (r arbitrary, s the smallest distance between O and r), always make that kind of angle.
e. Now you should be able - given any line r with equation 'y=mx+c' - to write an equation for s, and explicitly calculate their intersection, which is the point of least distance. I'm not sure what you mean by dot product, since there are many ways to express the result as a dot product. A possibly 'nice' way is hidden below.
Spoiler:
The answer turns out to be
|c| / [sqrt](m2+1)
which we might express as
|c| / [sqrt](v [cdot] v) or |c| / |v|
where v is the given line's direction vector, i.e. (m,1). I'm not sure why we should consider this direction vector canonical, though - (1,1/m) is just as nice.
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Rego
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thank you
its much clearer now 
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