Vectors

**Clairvoyantski** · June 11th 2012, 04:32 PM

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 ?

**HallsofIvy** · June 11th 2012, 06:52 PM

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).

**Soroban** · June 11th 2012, 08:51 PM

Hello, Clairvoyantski!

I have a rather roundabout solution . . .

$\text{There are two vectors, }\vec c\,\text{ and }\,\vec d. \;\;\;\vec d \cdot \vec c \,=\,50.$
$\text{The angle between them is }60^o\text{ (when the two vectors are tail to tail).}$

$\text{What is }|\vec d \times \vec c|\,?$

Code:

          C * - - - - - * B
           / \         /
          /   \       /
       c /     \     /
        /       \   /
       / 60d     \ /
    A * - - - - - * D
            d

We have parallelogram $AC\!BD,$
. . where $AD = \vec d,\;\;AC = \vec c.$

We know that $|\vec d \times \vec c|$ is the area of the parallelogram.

Formula: . $\cos60^o \:=\:\frac{\vec d \cdot \vec c}{|\vec d||\vec c|}$

So we have: . $\frac{1}{2} \:=\:\frac{50}{|\vec d||\vec c|} \quad\Rightarrow\quad |\vec d||\vec c| \:=\:100$

The area of $\Delta AC\!D$ is given by: . $\tfrac{1}{2}|\vec d||\vec c|\sin60^o$

The area of parallelogram $AC\!BD$ is: . $|\vec d||\vec c|\sin60^o \;=\;(100)\left(\tfrac{\sqrt{3}}{2}\right) \;=\;50\sqrt{3}$

Therefore: . $|\vec d \times \vec c| \;=\;50\sqrt{3}$

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