<PRE><B>Question 229691</B>
&lt;pre&gt;&lt;font size = 4 color = &quot;indigo&quot;&gt;
First we write it as a difference of vectors, 
but not necessarily unit vectors.

&amp;#10216;-7-(-1),5-3,1-(-2)&amp;#10217; =

&amp;#10216;-7+1,2,1+2&amp;#10217; =

&amp;#10216;-6,2,3&amp;#10217; =

-6&lt;b&gt;i&lt;/b&gt; + 2&lt;b&gt;j&lt;/b&gt; + 3&lt;b&gt;k&lt;/b&gt;

Now that is a sum of scalar multiples of unit vectors

&lt;b&gt;i&lt;/b&gt;, &lt;b&gt;j&lt;/b&gt;, and &lt;b&gt;k&lt;/b&gt;

I have no idea why you would want to write that as

a sum of unit vectors.  But since that&#39;s what you say

you want, since &lt;b&gt;i&lt;/b&gt;, &lt;b&gt;j&lt;/b&gt;, and &lt;b&gt;k&lt;/b&gt; are unit vectors, we could

write -6&lt;b&gt;i&lt;/b&gt; + 2&lt;b&gt;j&lt;/b&gt; + 3&lt;b&gt;k&lt;/b&gt;

as this sum of unit vectors:

-&lt;b&gt;i&lt;/b&gt;-&lt;b&gt;i&lt;/b&gt;-&lt;b&gt;i&lt;/b&gt;-&lt;b&gt;i&lt;/b&gt;-&lt;b&gt;i&lt;/b&gt;-&lt;b&gt;i&lt;/b&gt;+&lt;b&gt;j&lt;/b&gt;+&lt;b&gt;j&lt;/b&gt;+&lt;b&gt;k&lt;/b&gt;+&lt;b&gt;k&lt;/b&gt;+&lt;b&gt;k&lt;/b&gt;

but that is very silly.  Are you sure that&#39;s what

you were asked to do?  

Edwin&lt;/pre&gt;


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