Question 582498
<pre>
ax + by + cz + d = 0

Let's find two points in the plane V.

Let y = 0 and z = 0

ax + b(0) + c(0) + d = 0

              ax + d = 0
                   
                  ax = {{{-d/a}}}

So (-d/a,0,0) is one point in the plane V

Let x = 0 and z = 0

a(0) + by + c(0) + d = 0

              by + d = 0
                   
                  by = {{{-d/b}}}

So (0,{{{-d/b}}},0) is another point in the plane V

The vector between these two points is

&#10216;0-({{{-d/a}}}),{{{-d/b}}}-0,0-0&#10217;  =  &#10216;{{{d/a}}},{{{-d/b}}},0&#10217;

This vector &#10216;{{{d/a}}},{{{-d/b}}},0&#10217; is parallel to the plane V.

Let's find the dot product of it with &#10216;a,b,c&#10217;

&#10216;{{{d/a}}},{{{-d/b}}},0&#10217;•&#10216;a,b,c&#10217; = ({{{d/a}}})a + ({{{-d/b}}})b + 0(c) = d-d+0 = 0

Since that dot product is 0, the vector &#10216;a,b,c&#10217; is perpendicular
to to a vector parallel to the plane V and therefore is perpendicular
to the plane V.

Therefore &#10216;a,b,c&#10217; is normal to the plane V since its dot product
with a vector parallel to the plane V is 0.

The correct choice is A)

Edwin</pre>