Difference between revisions of "031 Review Part 1, Problem 8"

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!Solution:    
 
!Solution:    
 
|-
 
|-
|First, we switch to the limit to <math style="vertical-align: 0px">x</math> so that we can use L'Hopital's rule.
+
|Since &nbsp;<math style="vertical-align: -4px">\vec{v}\in W^\perp,</math>&nbsp; we know &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; is orthogonal to every vector in &nbsp;<math style="vertical-align: 0px">W.</math>
 
|-
 
|-
|So, we have
+
|In particular, since &nbsp;<math style="vertical-align: -4px">\vec{v}\in W,</math>&nbsp; we have that &nbsp;<math style="vertical-align: 0px">\vec{v}</math>&nbsp; is orthogonal to &nbsp;<math style="vertical-align: 0px">\vec{v}.</math>
 +
|-
 +
|Hence,
 
|-
 
|-
 
|
 
|
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
+
::<math>\vec{v}\cdot \vec{v}=0.</math>
\displaystyle{\lim_{x \rightarrow \infty}\frac{3-2x^2}{5x^2 + x +1}} & \overset{L'H}{=} & \displaystyle{\lim_{x \rightarrow \infty}\frac{-4x}{10x+1}}\\
+
|-
&&\\
+
|But, this tells us that &nbsp;<math style="vertical-align: 0px">\vec{v}=\vec{0}.</math>
& \overset{L'H}{=} & \displaystyle{\frac{-4}{10}}\\
+
|-
&&\\
+
|Therefore, the statement is true.
& = & \displaystyle{-\frac{2}{5}}.
 
\end{array}</math>
 
 
|}
 
|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|&nbsp;&nbsp; &nbsp; &nbsp; False
+
|&nbsp;&nbsp; &nbsp; &nbsp; TRUE
 
|}
 
|}
 
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]
 
[[031_Review_Part_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 14:22, 9 October 2017

True or false: Let    be a subspace of    and    be a vector in    If    and    then  

Solution:  
Since    we know    is orthogonal to every vector in  
In particular, since    we have that    is orthogonal to  
Hence,
But, this tells us that  
Therefore, the statement is true.
Final Answer:  
       TRUE

Return to Sample Exam