S6.793.12.4.27area of a parallelogram

[karush]

$\tiny{s6.793.12.4.27}$
$\textsf{area of a parallelogram with adjacent sides}\\$
$\textsf{$\vec{P}{Q}$ and $\vec{P}{R}$ is the length of this cross product:}$

\begin{align}
|PQ \times PR|&=\sqrt{()^2+()^2}=
\end{align}

$\textit{new on vectors...
not sure what goes into $()^2$}$

 

[HallsofIvy]

Are you sure it isn't "$$\sqrt{()^2+ ()^2+ ()^2}$$? The cross product of two vectors, in $$R^3$$, is a vector in $$R^3$$ of the form $$x\vec{i}+ y\vec{j}+ z\vec{k}$$ and its length is $$\sqrt{x^2+ y^2+ z^2}$$.

 

[karush]

$\tiny{s6.12.4.27}$
The points given are $(x, y)$ no $z$ values
$A(-2,1), B(0,4), C(4,2), D(2,-1)$
\begin{align}
\vec{A}{B}&=(0+2)i+(4-1)j=2i+3j\\
\vec{B}{C}&=(4-0)i+(2-4)j=4i-2j\\
\vec{A}{B} \times\vec{B}{C}&=
\left|\begin{bmatrix}
4 & -0 \\
2 & -4 \\
\end{bmatrix}\right|=\left|4(4)-0(2)\right|
=\color{red}{16}
\end{align}

 

[HallsofIvy]

$\tiny{s6.12.4.27}$
The points given are $(x, y)$ no $z$ values
$A(-2,1), B(0,4), C(4,2), D(2,-1)$
\begin{align}
\vec{A}{B}&=(0+2)i+(4-1)j=2i+3j\\
\vec{B}{C}&=(4-0)i+(2-4)j=4i-2j\\
\vec{A}{B} \times\vec{B}{C}&=
\left|\begin{bmatrix}
4 & -0 \\
2 & -4 \\
\end{bmatrix}\right|=\left|4(4)-0(2)\right|
=\color{red}{16}
\end{align}

That is using what I would consider "short cut" notation- that I don't like! The cross product of vectors is only defined for three dimensional vectors. Given AB= (2, 3) and BC=(4, -2) (why was D given at all?) I would write them as AB= (2, 3, 0) and BC= (4, -2, 0). Of course, then the cross product is a vector in the z direction. Then the cross product is $$AB\times BC= \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k}\\ 2 & 3 & 0 \\ 4 & -2 & 0 \end{array}\right|= (0, 0, -16)$$

 

[karush]

strictly speaking yes ..