1. ## Vectors and area

Im having trouble with two problems

Find the unit vectors perpendicular to the plane determined by the three points (-1, 3, 0), (5,1,2), and (4,-3,1).
i think here you have to divide something by its magnitude but im not sure.

Find the area of the triangle with (3,2,1), (2,4,6), and (-1,2,5) as vertices.
i can't even begin here!

any suggestions appreciated

2. For the first problem, you need to find a normal vector that defines the plane. To do this find two vectors in the plane using those points and cross them:

$A = (5, 1, 2) - (-1, 3, 0) = <6, -2, 0>$
$B = (4, -3, 1) - (-1, 3, 0) = <5, -6, 0>$

$A X B = <0, 0, -26> = C$

Now this is the normal vector that defines the plane, we need to get the unit vector:

Unit Vector: $\frac{C}{|C|} = \frac{<0, 0, -26>}{26} = <0, 0, -1>$ also $<0, 0, 1>$

We must include both the negative and positive of this unit vector because it is perpendicular either way. I hope that's pretty clear, if not, I'll explain more just ask.

For the second problem, take the absolute value of the determinant divided by 2:

$\left(\begin{array}{cccc}3&2&1\\2&4&6\\-1&2&5\end{array}\right) = 0$

I believe these points to be co-linear since I calculated the determinant to be 0, but I'm going to check again. So for now, I believe the area of the triangle to be 0.

3. Originally Posted by eXist
For the first problem, you need to find a normal vector that defines the plane. To do this find two vectors in the plane using those points and cross them:

$A = (5, 1, 2) - (-1, 3, 0) = <6, -2, 0>$
$B = (4, -3, 1) - (-1, 3, 0) = <5, -6, 0>$

$A X B = <0, 0, -26> = C$

Now this is the normal vector that defines the plane, we need to get the unit vector:

Unit Vector: $\frac{C}{|C|} = \frac{<0, 0, -26>}{26} = <0, 0, -1>$ also $<0, 0, 1>$

We must include both the negative and positive of this unit vector because it is perpendicular either way. I hope that's pretty clear, if not, I'll explain more just ask.

For the second problem, take the absolute value of the determinant divided by 2:

$\left(\begin{array}{cccc}3&2&1\\2&4&6\\-1&2&5\end{array}\right) = 0$

I believe these points to be co-linear since I calculated the determinant to be 0, but I'm going to check again. So for now, I believe the area of the triangle to be 0.
No, the points are not collinear but they are co-planar, of course- these are 3 points, not vectors, and 3 points always lie in one plane. The "triple" product $\vec{u}\cdot(\vec{v}\times\vec{w})$, of three vectors in $R^3$ gives the volume of the parallelopiped having those three vectors as edges.
The formula you need is the two dimensional equivalent: the area of the parallelogram having $\vec{u}$ and $\vec{v}$ as sides is the length of the cross product of the sides. Given the points (3,2,1), (2,4,6), and (-1,2,5), form the vectors $(2-3)\vec{i}+ (4-2)\vec{j}+ (6- 1)\vec{k}$ and $(-1-3)\vec{i}+ (2-2)\vec{j}+ (5-1)\vec{k}$. The length of their cross product is the area of a parallelogram having those vectors as sides and 1/2 that is the area of the triangle.