<PRE><B>Question 196181</B>
&lt;pre&gt;&lt;font size = 4 color = &quot;indigo&quot;&gt;&lt;b&gt;

Italicized letters such as &lt;i&gt;a&lt;/i&gt; represent vectors
Regular letters represent scalars which are
just plain old numbers.

The scalar product:

That&#39;s the kind of multiplying where you multiply
two vectors and get a &quot;non-vector&quot;, called a scalar,
which is nothing but a PLAIN OLD NUMBER.

There are two separate formulas for the scalar product.
You use whichever one you need depending on what you
have given.

The first formula is the basic one:

&lt;i&gt;a&lt;/i&gt; = m&lt;i&gt;i&lt;/i&gt; + n&lt;i&gt;j&lt;/i&gt; + p&lt;i&gt;k&lt;/i&gt; 
&lt;i&gt;b&lt;/i&gt; = q&lt;i&gt;i&lt;/i&gt; + r&lt;i&gt;j&lt;/i&gt; + s&lt;i&gt;k&lt;/i&gt;

&lt;i&gt;a&lt;/i&gt;·&lt;i&gt;b&lt;/i&gt; = (m&lt;i&gt;i&lt;/i&gt; + n&lt;i&gt;j&lt;/i&gt; + p&lt;i&gt;k&lt;/i&gt;)·(q&lt;i&gt;i&lt;/i&gt; + r&lt;i&gt;j&lt;/i&gt; + s&lt;i&gt;k&lt;/i&gt;) = mq + nr + ps

so for

(2&lt;i&gt;i&lt;/i&gt; - &lt;i&gt;j&lt;/i&gt; + 3&lt;i&gt;k&lt;/i&gt;)·(-&lt;i&gt;i&lt;/i&gt; - 7&lt;i&gt;j&lt;/i&gt;)

First we rewrite it to show the coefficients
-1 on &lt;i&gt;j&lt;/i&gt; in the first and &lt;i&gt;i&lt;/i&gt; in the second, and 
put &quot;+ 0&lt;i&gt;k&lt;/i&gt;&quot; as a placeholder for the &lt;i&gt;k&lt;/i&gt;-component 
in the second one:

(2&lt;i&gt;i&lt;/i&gt; - 1&lt;i&gt;j&lt;/i&gt; + 3&lt;i&gt;k&lt;/i&gt;)·(-1&lt;i&gt;i&lt;/i&gt; - 7&lt;i&gt;j&lt;/i&gt; + 0&lt;i&gt;k&lt;/i&gt;)

m=2, n=-1, p=3, q=-1, r=-7, s=0

So

(m&lt;i&gt;i&lt;/i&gt; + n&lt;i&gt;j&lt;/i&gt; + p&lt;i&gt;k&lt;/i&gt;)·(q&lt;i&gt;i&lt;/i&gt; + r&lt;i&gt;j&lt;/i&gt; + s&lt;i&gt;k&lt;/i&gt;) = mq + nr + ps

becomes

[(2)&lt;i&gt;i&lt;/i&gt; + (-1)&lt;i&gt;j&lt;/i&gt; + (3)&lt;i&gt;k&lt;/i&gt;]·[(-1)&lt;i&gt;i&lt;/i&gt; + (-7)&lt;i&gt;j&lt;/i&gt; + (0)&lt;i&gt;i&lt;/i&gt;] =

(2)(-1) + (-1)(-7) + (3)(0) = -2 + 7 + 0 = 5.

So their scalar product (also called &quot;dot product&quot;)
is simply the scalar, (or plain old number), 5.
 
Now we look at the other formula for the scalar product:

&lt;i&gt;a&lt;/i&gt;·&lt;i&gt;b&lt;/i&gt; = |&lt;i&gt;a&lt;/i&gt;||&lt;i&gt;b&lt;/i&gt;|cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt; where &lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt; is the angle between the 

vectors &lt;i&gt;a&lt;/i&gt; and &lt;i&gt;b&lt;/i&gt;, and where |&lt;i&gt;a&lt;/i&gt;| and |&lt;i&gt;b&lt;/i&gt;| represent the
scalar (plain old number) magnitude, which is just
the length of the vector. It is found by the extended
Pythagorean formula:

|&lt;i&gt;a&lt;/i&gt;| = |m&lt;i&gt;i&lt;/i&gt; + n&lt;i&gt;j&lt;/i&gt; + p&lt;i&gt;k&lt;/i&gt;| = 

&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(m&lt;sup&gt;2&lt;/sup&gt; + n&lt;sup&gt;2&lt;/sup&gt; + p&lt;sup&gt;2&lt;/sup&gt;) 
|&lt;i&gt;b&lt;/i&gt;| =

|q&lt;i&gt;i&lt;/i&gt; + r&lt;i&gt;j&lt;/i&gt; + s&lt;i&gt;k&lt;/i&gt;| =
 ____________ 
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(q&lt;sup&gt;2&lt;/sup&gt; + r&lt;sup&gt;2&lt;/sup&gt; + s&lt;sup&gt;2&lt;/sup&gt;)

So:

|&lt;i&gt;a&lt;/i&gt;| = |2&lt;i&gt;i&lt;/i&gt; - 1&lt;i&gt;j&lt;/i&gt; + 3&lt;i&gt;k&lt;/i&gt;| = 
 ___________________
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(2)&lt;sup&gt;2&lt;/sup&gt; + (-1)&lt;sup&gt;2&lt;/sup&gt; + (3)&lt;sup&gt;2&lt;/sup&gt; =
 _________
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;4 + 1 + 9 = 
 __
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;14

and
 
|&lt;i&gt;b&lt;/i&gt;| = |-1&lt;i&gt;i&lt;/i&gt; - 7&lt;i&gt;j&lt;/i&gt; + 0&lt;i&gt;k&lt;/i&gt;| =
 ___________________
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(-1)&lt;sup&gt;2&lt;/sup&gt; + (-7)&lt;sup&gt;2&lt;/sup&gt; + (0)&lt;sup&gt;2&lt;/sup&gt; =
 ____________
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;1 + (49) + 0 = 
 __                                 
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;50 =
  _______
&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(25)(2) = 
  _
5&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;2

Substituting in 

&lt;i&gt;a&lt;/i&gt;·&lt;i&gt;b&lt;/i&gt; = |&lt;i&gt;a&lt;/i&gt;||&lt;i&gt;b&lt;/i&gt;|cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;
      __                                 _
5 = (&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;14)(5&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;2)cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;
      __
5 = 5&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;28)cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;
      ______
5 = 5&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(4)(7)*cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;
         _
5 = 5(2)&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;7*cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;
       _                
5 = 10&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;7*cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;
        
{{{5/(10sqrt(7))}}} = cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;


{{{1/(2sqrt(7))}}} = cos&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt;

Get the inverse cosine of {{{1/(2sqrt(7))}}}

&lt;font face = &quot;symbol&quot;&gt;q&lt;/font&gt; = 79.10° approximately

Edwin&lt;/pre&gt;
</PRE>