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\centerline
{
\bf Dr. Z's Math251 Handout \#12.4 (2nd ed.) [The Cross Product]
}

By Doron Zeilberger

{\bf Problem Type 12.4a}: 
Find the cross product
${\bf a} \times {\bf b}$ and verify that it is
orthogonal to both ${\bf a}$ and ${\bf b}$.
$$
{\bf a}= \langle a_1,a_2,a_3 \rangle \quad , \quad 
{\bf b}=\langle b_1,b_2,b_3 \rangle \quad.
$$
%%General Problem

{\bf Example Problem 12.4a}: 
Find the cross product
${\bf a} \times {\bf b}$ and verify that it is
orthogonal to both ${\bf a}$ and ${\bf b}$.
$$
{\bf a}=\langle 4,5,-3 \rangle \quad , \quad {\bf b}=
\langle 2,1,-3 \rangle \quad.
$$

%%Specific Problem
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{\bf Steps} & {\bf Example} \cr  &  \cr

{\bf 1.} 
%%%General Step 1
Form the {\bf determinant} with
${\bf i},{\bf j},{\bf k}$ in the first row,
the vector ${\bf a}$ in the second row and
the vector ${\bf b}$ in the third row.
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
a_1 & a_2 & a_3\cr
b_1 & b_2 & b_3 
} \right \vert \quad .
$$
%%%End General Step 1
&
{\bf 1.} 
%%%Specific Step 1
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
4 & 5& -3\cr
2 & 1 & -3 
} \right \vert \quad .
$$
%%%End Specific Step 1
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{\bf 2.} Evaluate the determinant step-by-step:
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
a_1 & a_2 & a_3\cr
b_1 & b_2 & b_3 
} \right \vert \quad =
$$
$$
{\bf i} \left \vert \matrix{ 
a_2 & a_3\cr
b_2 & b_3 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
a_1 & a_3\cr
b_1 & b_3 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
a_1 & a_2\cr
b_1 & b_2 
} \right \vert 
$$
$$
={\bf i} (a_2b_3-b_2a_3)
-{\bf j} (a_1b_3-b_1a_3)
+{\bf k} (a_1b_2-b_1a_2) \quad .
$$


%%% End General Step 2
&
{\bf 2.} 
%%%Specific Step 2
$$
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
4 & 5& -3\cr
2 & 1 & -3 
} \right \vert \quad=
$$
$$
{\bf i} \left \vert \matrix{ 
5 & -3\cr
1 & -3 
} \right \vert 
-{\bf j} \left \vert \matrix{ 
4 & -3\cr
2 & -3 
} \right \vert 
+{\bf k} \left \vert \matrix{ 
4 & 5 \cr
2 & 1 
} \right \vert 
$$
$$
={\bf i} (5 \cdot (-3)-(1) \cdot (-3))
-{\bf j} (4 \cdot (-3)-(2) \cdot (-3))
+{\bf k} (4 \cdot (1)-(2) \cdot (5))=
$$
$$
-12 {\bf i}+ 6 {\bf j} - 6 {\bf k}
$$

{\bf First Part of the Answer:}

$-12 {\bf i}+ 6 {\bf j} - 6 {\bf k}$ or
$\langle -12,6,-6 \rangle$.
%%%End Specific Step 2
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{\bf 3.} Take the dot product of this vector with 
${\bf a}$ and ${\bf b}$ and verify that they are
both $0$. If you get something else, something went
wrong.

&
{\bf 3.} 
$\langle -12,6,-6 \rangle. \langle 4,5,-3 \rangle=$
$(-12)(4)+(6)(5)+(-6)(-3)=-48+30+18=0$

and

$\langle -12,6,-6 \rangle. \langle 2,1,-3 \rangle=$
$(-12)(2)+(6)(1)+(-6)(-3)=-24+6+18=0$.

{\bf Second Part of Ans.}: Both are zero as expected.

%%% End Specific Step 3
\cr}

{\bf Problem Type 12.4b}: 
(a) Find a vector orthogonal
to the plane through the points $P=(p_1,p_2,p_3)$,
$Q=(q_1,q_2,q_3)$,$R=(r_1,r_2,r_3)$.
(b) Find the area of the triangle $PQR$.

{\bf Example Problem 12.4b}: 
(a) Find a vector orthogonal
to the plane through the points $P=(2,1,5)$,
$Q=(-1,3,4)$,$R=(3,0,6)$.
(b) Find the area of the triangle $PQR$.

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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.} Compute 
the vector  ${\bf PQ}$ by
subtracting $Q$ from $P$ and 
the vector  ${\bf PR}$ by
subtracting  $R$ from $P$.
&
{\bf 1.} 
%%%Specific Step 1
$${\bf PQ}=(-1,3,4)-(2,1,5)=
\langle -3,2,-1 \rangle \quad ,
$$
$${\bf PR}=(3,0,6)-(2,1,5)=
\langle 1,-1,1 \rangle \quad .
$$

%%%End Specific Step 1
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{\bf 2.} Take the cross product of ${\bf PQ}$
and ${\bf PR}$.

&

$$
{\bf PQ} \times {\bf PR}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
-3 & 2 & -1\cr
1 & -1 & 1 
} \right \vert \quad .
$$
$$
=
{\bf i} \left \vert \matrix{
2 & -1 \cr
-1 & 1  
} \right \vert
-{\bf j} \left \vert \matrix{
-3 & -1 \cr
1 & 1  
} \right \vert
+{\bf k} \left \vert \matrix{
-3 & 2 \cr
1 & -1  
} \right \vert
$$
$$
={\bf i}+2{\bf j}+{\bf k}=
\langle 1,2,1 \rangle \quad .
$$
{\bf Ans. to First Part:}
A vector  orthogonal to the
plane of the triangle $PQR$ is
$\langle 1,2,1 \rangle$.
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{\bf 3.} Take the magnitude of the vector you found in
step 2. This is the {\bf area of the parralelogram}.
To get the {\bf area of the triangle}, divide it by $2$.

&
{\bf 3.} 
$$
\vert \langle 1,2,1 \rangle \vert =\sqrt{1^2+2^2+1^2}=\sqrt{6} \quad.
$$
So the area of the parallelogram formed by $PQ$ and $PR$ is $\sqrt{6}$
and the area of triangle PQR is half of that, $\sqrt{6}/2$.

{\bf Ans. to Second Part:} The area of triangle
$PQR$ is $\sqrt{6}/2$.

\cr}

{\bf Problem Type 12.4c}: 
Find the volume of the
parallelopiped with adjacent edges
$PQ$,$PR$, and $PS$, where $P,Q,R,S$ are given points.

%%General Problem
{\bf Example Problem 12.4c}: 
Find the volume of the
parallelopiped with adjacent edges
$PQ$,$PR$, and $PS$, where 
$P(0,1,2),Q(2,4,5),R(-1,0,1),S(6,-1,4)$.

%%Specific Problem
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{\bf Steps} & {\bf Example} \cr  &  \cr
{\bf 1.}  Compute the vectors $\bf PQ$,
$\bf PR$,$\bf PS$ by the appropriate subtractions.
&
{\bf 1.} 
$$
{\bf PQ}=Q-P=(2,4,5)-(0,1,2)=\langle 2,3,3 \rangle \quad ,
$$
$$
{\bf PR}=R-P=(-1,0,1)-(0,1,2)=\langle -1,-1,-1 \rangle \quad ,
$$
$$
{\bf PS}=S-P=(6,-1,4)-(0,1,2)=\langle 6,-2,2 \rangle \quad .
$$

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{\bf 2.} Compute the cross product ${\bf PQ} \times {\bf PR}$.

%%% End General Step 2
&
{\bf 2.} 
$$
{\bf PQ} \times {\bf PR}=
\left \vert \matrix{ {\bf i} & {\bf j} & {\bf k}  \cr
2 & 3 & 3\cr
-1 & -1 & -1 
} \right \vert \quad .
$$
$$
=
{\bf i} \left \vert \matrix{
3 & 3 \cr
-1 & -1  
} \right \vert
-{\bf j} \left \vert \matrix{
2 & 3 \cr
-1 & -1  
} \right \vert
+{\bf k} \left \vert \matrix{
2 & 3 \cr
-1 & -1  
} \right \vert
$$
$$
=0 \cdot {\bf i}-(1) \cdot {\bf j}+(1) \cdot {\bf k}=
-{\bf j}+{\bf k}=
\langle 0,-1,1 \rangle
$$

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{\bf 3.} Compute the dot product of what you found in
step 2 with the vector ${\bf PS}$.
&
{\bf 3.} 
$$
({\bf PQ} \times {\bf PR}) \cdot {\bf PS}=
\langle 0,-1,1 \rangle .\langle 6,-2,2 \rangle=
$$
$$
(0)\cdot (6)+ (-1) \cdot (-2)+ (1) \cdot (2)=4 \quad .
$$

{\bf Ans.:} The volume of the parallelopiped is
$4$.

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\end