Question 674684
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A(1, 1, 0) B(2, 1, 1) and C(1, 1, 2)


The magnitude of the cross product of two vectors
is the area of the parallelogram which has adjacent
sides equal to the magnitudes of the two vectors.

We form the vectors <span style="text-decoration: overline">AB</span> and <span style="text-decoration: overline">AC</span>

<span style="text-decoration: overline">AB</span> = < 2-1, 1-1, 1-0 > = < 1, 0, 1 >

<span style="text-decoration: overline">AC</span> = < 1-1, 1-1, 2-0 > = < 0, 0, 2 >

We find their cross product <span style="text-decoration: overline">AB</span>×<span style="text-decoration: overline">AC</span>

{{{abs(matrix(3,3,i,j,k,1,0,1,0,0,2))}}} = [(0)(2)-(1)(0)]<span style="text-decoration: overline">i</span> - [(1)(2)-(1)(0)]<span style="text-decoration: overline">j</span> + [(1)(0)-(0)(0)]<span style="text-decoration: overline">k</span>
 
= 0<span style="text-decoration: overline">i</span> - (2-0)<span style="text-decoration: overline">j</span> + 0<span style="text-decoration: overline">k</span> = < 0, -2, 0 >

The magnitude of < 0, -2, 0 > is
 
&#8730;<span style="text-decoration: overline">0² + (-2)² + 0²</span> = &#8730;<span style="text-decoration: overline">0 + 4 + 0</span> = &#8730;<span style="text-decoration: overline">4</span> = 2

The triangle's area is {{{1/2}}} the area of that parallelogram.

Answer = {{{1/2}}}(2) = 1

Edwin</pre>