Recent content by simpleas123
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Mechanics with vectors question
Alright all I've been attempting this question and I've become stuck. This is the question. Homework Statement http://img707.imageshack.us/img707/2575/63353658.jpg Homework Equations In the working out. The Attempt at a Solution So for the first part i wrote F=Mx^.. (X...- simpleas123
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- Mechanics Vectors
- Replies: 1
- Forum: Introductory Physics Homework Help
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Need help with this Vector Question +rep
Oh snap that's right, well then how would i prove it?- simpleas123
- Post #10
- Forum: Calculus and Beyond Homework Help
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Need help with this Vector Question +rep
Part ii) c \cdot y = c \cdot x (c \cdot y) - (c \cdot x) = 0 c \cdot (x-y) = 0 \therefore y = x and it was pretty much the same method for the cross product execpt u had to state a property at the end as x x y could equal something other than 0.- simpleas123
- Post #7
- Forum: Calculus and Beyond Homework Help
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Need help with this Vector Question +rep
Property 4 is the same as result of problem 3 (b). It states that; Where a,b are vectors a x b = 0, if and only if there exist two scalars \mu,\lambda one of which is non-zero such that a\lambda = b\mu I have done part ii) now too.- simpleas123
- Post #5
- Forum: Calculus and Beyond Homework Help
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Need help with this Vector Question +rep
c = a + t(b-a), take t = l1 / l c = a + (l1 / l)(b-a), Expand c = a -a(l1 / l) + b(l1 / l) Collect terms with a. c = a(1 - (l1 / l)) + b(l1 / l) 1 = l / l c = a(l/l - (l1 / l)) + b(l1 / l) c = a(l - l1)/l + b(l1 / l) l - l1 = l2 c = a(l2 / l) + b(l1 / l) Its the rest I...- simpleas123
- Post #3
- Forum: Calculus and Beyond Homework Help
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Need help with this Vector Question +rep
Alright all I am on this vector question and I am having problems with this question. Part (i) i have no idea what to start with / how i possibly prove the question. This is the question, help will be deeply appriciated. Thanks, David. Also note that || a || = (a_1 + a_2 + ... + 1_n)^1/2 ...- simpleas123
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- Vector
- Replies: 10
- Forum: Calculus and Beyond Homework Help