What Is the Dimension of Sp{v1,..,vk,u,w}?

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The discussion focuses on determining the dimension of the span of the vectors {v1, ..., vk, u, w}, given that {v1, ..., vk, u} is independent and the intersection of the spans of {v1, ..., vk} and {u, w} is non-trivial. The dimension of the span of {v1, ..., vk} is k, and since {v1, ..., vk, u} is independent, the dimension of its span is k + 1. The formula for the dimension of the combined span is provided, which accounts for the intersection of the two spans. The independence of u from {v1, ..., vk} suggests that the intersection of their spans is trivial, simplifying the calculation. Ultimately, the dimension of sp{v1, ..., vk, u, w} can be derived from these relationships.
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there is
v1,..,vk,u,w vectors on space V
the group {v1,..,vk,u} is independant
and
Sp\{v1,..,vk\}\cap Sp\{u,w\}\neq\{0\}
find the dimention of sp{v1..vk,u,w}
?

if {v1,..,vk,u} is independant then its span dimention is k+1
and span so {v1,..,vk} independat to and dim sp{v1,..,vk}=k

the dimention of sp{v1..vk,u,w} is
dim(sp{v1..vk,u,w})=dim(sp{v1..vk}and sp{u,w}})=dim(sp{v1..vk}+sp{u,w})=dim(sp{v1..vk})+dim(sp{u,w})-dim(sp{v1..vk} intersect sp{u,w}})

if v1..vk,u is independant then v1,..,vk is independant too so dim (sp{v1..vk})=k

now regardingthe other two members i don't know
 
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You are given that u is indendent of v1,v2,...,vk.
What does this imply about the intersection of the sapces spanned by {v1,v2...,vk} and {u}?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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