Area of triangle formed by 2 vectors

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Homework Statement


I have a triangle ABC formed by vectors a = ( 2, 10, 25 ) and b = ( 5, 4, -2 ) attached at a point P.
And I have to calculate the area of the triangle ABC.


Homework Equations


I'm pretty certain that I need to find the area of the parellogram and then half it.
So I need to find the cross product, by calculating the determinant.


The Attempt at a Solution


Calculating the determinant I got
-120 i + 129 j - 42 k
so does this mean i square those and take the square root of them to find the area of the parellogram and then half it...
= 181.121
=> area of triangle = 90.56
 
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Sure, you're doing it right.
 
Thanks..I thought I was on the right lines, just needed reassuring :)

I got the area of the traingle to be 81root5 / 2

But now I need to calculate the length of the height PM (where PM perpendicular to BC and the point M belongs to the line BC)

I'm guessing as I know the area and 2 vectors I can take it from there...but I really have no idea...
 
Use |axb|=|a|*|b|*sin(t), where t is the smaller angle between the vectors a and b. Find t and use trig.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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