<PRE><B>Question 54307</B>
&lt;pre&gt;&lt;font size = 5&gt;find the distance from point Q to line l (matrix)

Q=(0, 1, 0)

line = &lt; x y z &gt; = &lt; 1 1 1 &gt; + t&lt; -2 0 3 &gt;

Formula:
       __
     ||PQ × &lt;b&gt;u&lt;/b&gt;||
D = ------------
       ||&lt;b&gt;u&lt;/b&gt;||

where &lt;b&gt;u&lt;/b&gt; is the direction vector for the line and
P is a point on the line

Using the direction numbers -2, 0, and 4, you know
that the direction vector for the line is

&lt;b&gt;u&lt;/b&gt; = &lt; -2, 0, 4 &gt;

To find a point on the line, let t = 0 and obtain

P = (1, 1, 1)

Thus, since Q = (0, 1, 0)
__
PQ = &lt; 0-1, 1-1, 0-1 &gt; = &lt; -1,0,-1 &gt;
                          __
Now get the cross product PQ × &lt;b&gt;u&lt;/b&gt;

__       | &lt;b&gt;i&lt;/b&gt;  &lt;b&gt;j&lt;/b&gt;  &lt;b&gt;k&lt;/b&gt;| 
PQ × &lt;b&gt;u&lt;/b&gt; = |-1  0 -1| = &lt;b&gt;i&lt;/b&gt;(0+0) - &lt;b&gt;j&lt;/b&gt;(-4-2) + &lt;b&gt;k&lt;/b&gt;(0+0) = -6&lt;b&gt;j&lt;/b&gt;
         |-2  0  4|
  __
||PQ × &lt;b&gt;u&lt;/b&gt;|| = ||-6&lt;b&gt;j&lt;/b&gt;|| = 6
         ____________    ______    __     _
||&lt;b&gt;u&lt;/b&gt;|| = &lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;(-2)²+0²+ 4² = &lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;4+0+16 = &lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;20 = 2&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;5
       __                            _
     ||PQ × &lt;b&gt;u&lt;/b&gt;||      6       3     3&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;5
D = ------------ = ---&lt;u&gt;-&lt;/u&gt;- = --&lt;u&gt;-&lt;/u&gt;- = -----
       ||&lt;b&gt;u&lt;/b&gt;||        2&lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;5     &lt;font face = &quot;symbol&quot;&gt;Ö&lt;/font&gt;5      5

Edwin&lt;/pre&gt;
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