+0  
 
0
138
1
avatar+956 

I got : 

 

Sorry if this is a really obvious problem, I'm just insecure with Vectors. 

Julius  May 3, 2018

Best Answer 

 #1
avatar+20147 
+2

Did I do it correct?

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{i} + 2\vec{k} \\ \vec{y} &=& 3\vec{i} + \vec{j} \\ \vec{z} &=& 3\vec{j} + \vec{k} \\\\ && \mathbf{2(\vec{x}-\vec{y})+\vec{z}} \\ &=& 2\left(\vec{i} + 2\vec{k}-(3\vec{i} + \vec{j}) \right)+3\vec{j} + \vec{k} \\ &=& 2\left(\vec{i} + 2\vec{k}-3\vec{i} - \vec{j} \right)+3\vec{j} + \vec{k} \\ &=& 2\left( -2\vec{i} - \vec{j} +2\vec{k} \right)+3\vec{j} + \vec{k} \\ &=& -4\vec{i} - 2\vec{j} +4\vec{k} +3\vec{j} + \vec{k} \\ &\mathbf{=}& \mathbf{-4\vec{i} +\vec{j} + 5\vec{k}} \\ \hline \end{array}\)

 

 

laugh

heureka  May 3, 2018
 #1
avatar+20147 
+2
Best Answer

Did I do it correct?

\(\begin{array}{|rcll|} \hline \vec{x} &=& \vec{i} + 2\vec{k} \\ \vec{y} &=& 3\vec{i} + \vec{j} \\ \vec{z} &=& 3\vec{j} + \vec{k} \\\\ && \mathbf{2(\vec{x}-\vec{y})+\vec{z}} \\ &=& 2\left(\vec{i} + 2\vec{k}-(3\vec{i} + \vec{j}) \right)+3\vec{j} + \vec{k} \\ &=& 2\left(\vec{i} + 2\vec{k}-3\vec{i} - \vec{j} \right)+3\vec{j} + \vec{k} \\ &=& 2\left( -2\vec{i} - \vec{j} +2\vec{k} \right)+3\vec{j} + \vec{k} \\ &=& -4\vec{i} - 2\vec{j} +4\vec{k} +3\vec{j} + \vec{k} \\ &\mathbf{=}& \mathbf{-4\vec{i} +\vec{j} + 5\vec{k}} \\ \hline \end{array}\)

 

 

laugh

heureka  May 3, 2018

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