There are two vectors (d and c). d dot c will equal 50. The angle between them is 60 degrees (when the two vectors are tail to tail). What is ID X CI ?
In order to do this equation, you need to know that d dot c= |d||c|cos(theta) where theta is the angle between them. I started to write "to find ID X CI" you would have to know what 'ID' and 'CI' mean" when it hit me that you mean |D X C|, the length of DXC. Of course, to do that you have to know that that length is given by |D||C| sin(theta).
Hello, Clairvoyantski!
I have a rather roundabout solution . . .
$\displaystyle \text{There are two vectors, }\vec c\,\text{ and }\,\vec d. \;\;\;\vec d \cdot \vec c \,=\,50.$
$\displaystyle \text{The angle between them is }60^o\text{ (when the two vectors are tail to tail).}$
$\displaystyle \text{What is }|\vec d \times \vec c|\,?$
We have parallelogram $\displaystyle AC\!BD,$Code:C * - - - - - * B / \ / / \ / c / \ / / \ / / 60d \ / A * - - - - - * D d
. . where $\displaystyle AD = \vec d,\;\;AC = \vec c.$
We know that $\displaystyle |\vec d \times \vec c|$ is the area of the parallelogram.
Formula: .$\displaystyle \cos60^o \:=\:\frac{\vec d \cdot \vec c}{|\vec d||\vec c|} $
So we have: .$\displaystyle \frac{1}{2} \:=\:\frac{50}{|\vec d||\vec c|} \quad\Rightarrow\quad |\vec d||\vec c| \:=\:100 $
The area of $\displaystyle \Delta AC\!D$ is given by: .$\displaystyle \tfrac{1}{2}|\vec d||\vec c|\sin60^o$
The area of parallelogram $\displaystyle AC\!BD$ is: .$\displaystyle |\vec d||\vec c|\sin60^o \;=\;(100)\left(\tfrac{\sqrt{3}}{2}\right) \;=\;50\sqrt{3}$
Therefore: .$\displaystyle |\vec d \times \vec c| \;=\;50\sqrt{3}$