<p>why are there more than 2 numbers in the coordinate..isn't it supposed to be (x,y) ?? am i missing something?</p>
<p>obsessedAndre - sorry, it was me who made a mistake somewhere (just as I was dreading). :o
adimeola and Gospy are right - were we allowed to use the cross product, getting the correct answer (~7.65) would be a breeze.<br>
Not so in Algebra 2. :mad:</p>
<p>bio_freak - are you sure you were in the right class? :)</p>
<p>dudes..... its soooo much easier then your making it out to be, enclose the non right triangle in a rectangle, then find the area of the rectangle and subtract from the two nice and even right triangles, no distance formula or any of that crap needed.</p>
<p>aputzer613... that doesn't make the problem any shorter, as you still have to find the dimensions of the rectangle. One dimension is easy to find but finding the other dimension is still somewhat lengthy. (Not terribly so.) And by the way, you would still need the distance formula...How would you know any of the dimensions of the rectangle otherwise?</p>
<p>As far as how the vector solution works...
Say you have a parallelogram, ABCD. (A is the bottom left vertice, B the top left, C the top right, D the bottom right.) Once again, drawing this helps while reading through the following. The area of this parallelogram is base times the height, bh. So, let us consider AD as the base of the parallelogram, so the length of the base is equal to the magnitude of AD. Now, to find the height. Say angle theta is the angle formed between AB and AD. The sin of this angle multiplied by the magnitude of AB is equal to the height of the parallelogram. At the same time,(a theorem of vector math) t the sin of an angle formed between two vectors is equal to the magnitude of the cross product of those vectors divided by the product of the magnitudes of those vectors. So if we have a vector in the direction of AB as i, and a vector in the direction of AD as j, then:
sin(theta) = ||i x j||/(||i||<em>||j||)
Therefore, according to the previous discussion, the area of the parallelogram:
= bh
= ||j||</em>||i||<em>||i x j||/(||i||</em>||j||) <------ remember, i = AB, j = AD.
= ||i X j||</p>
<p>So the area of a parallelogram defined by two vectors i, j corresponding to two sides of the parallelogram is given by the magnitude of the crosee product of those vectors.</p>
<p>Since a triangle is simply half of a parallelogram (for example, in your sketch draw in the diagnol of ABCD, youll see what I mean.) the area is 1/2 the cross product of the vectors.</p>
<p>lol........</p>
<p>I'm trying to make sense of what adimeola wrote, and while I think I understand it, I'm not quite sure I'm doing it right. (I've never worked with vectors before so bare with me)</p>
<p>I drew the parallelogram as you described it and understand the trigonometric relationship. Now, I'm trying to get the magnitude of the cross product.</p>
<p>The cross product is defined as a x b = |a||b|sin t, where t is the angle included by sides AB and AD of the paralellogram. In the original triangle biofreak posted, there was no AD, so I'm assuming we would have to find angle theta included between sides AB and BC, and then find it's supplementary angle, to get angle t, correct?</p>
<p>This is where I'm stuck. How do you find the angle theta included betewen sides AB and BC? I tried a formula for cos t = |A|^2+|B|^2 blah blah but I got t = 0 so I must have made an error somewhere.</p>
<p>ObsessedAndre:
Shouldn't your formula be
cos( 180 - t) = [ (AB)^2 + (AC)^2 - (BC)^2 ] / [ 2 (AB)(AC) ] ?</p>
<p>Plug in the numbers for the sides AB,BC,AC and see if it makes sense.</p>
<p>Using your AB=sqrt(27), AC=sqrt(17), BC=sqrt(14), I get
cos(180 - t) = 0.7001
or 180 - t = 45.56 degrees
so t = 134.44 degrees (assuming no math errors above)</p>
<p>I get those same values using the formula you described. </p>
<p>Now if we compute sin(t)*|a||b| we should get</p>
<p>sin(134.44 degrees)(sqrt27)(sqrt14) using |a| = length of AB and |b| = length of BC</p>
<p>which gives 13.881 as the area of the parallelogram; therefore the area of the triangle as 13.881/2 = 6.94.</p>
<p>But this is different from gcf101's answer of 6.5, so I'm not sure if it's right?</p>
<p>I was thinking about this one and here's what I got:
Forget about Heron's formula, as I've stated, I forgot it could be done only with integers.
But, why just not graph the triangle on the xyz plane, and find the base and height...
Base would be obvious how to calculate, and height could be calculated by using a right triangle "behind" the graphed triangle.</p>
<p>
[quote]
But this is different from gcf101's answer of 6.5
[/quote]
obsessedAndre -
I think your first answer in post #15 was correct (see my post #22).</p>
<p>This question is one of the calculus most commonly used illustrations of using the cross product for determining the area of a triangle in R3. The easiest way to do it is through the determinants. Plain pen/paper/calculator activity.
I was hoping there is something special about the coordinates that could lead to a shortcut. Seems we ain't finding no stinking shortcut.
This question is far beyond either Algebra 2 or SAT.
I suggest we end this thread as unrelated to developing either problem solving or SAT taking skills.</p>
<p>damn i hope none of my questions on my SAT are going to be this hard.</p>