Expert: Steve Holleran - 4/18/2007

Question
Hi im in my 1st year of University doing an introduction to algebra paper. I was just wondering if you could check my answers for me in one question and help me out with another. Any help appreciated.

1. let u=2i-3j+4k, v=i-j+2k. Find(using the cross product):
(a) a vector perpendicular to both u & v
---- i get (-2,0,1) is that right?
(b) the area of the parallelogram defined by u and v
---- i get sqrt(5) units
(c) the sine of the angle between u and v
---- have many different answers, just can't get this one


2. Consider the triangle in 3-space which has as vertices the points P,Q,R with respective position vecotrs of (1,0,1),(-1,1,0) and (2,1,2). Find vectors u and v describing the sides of the triangle which meet at P. Use a vector product to work out the angle at P, and thus calculate the area of the triangle. Use a similar method to calculate the angles at Q and R. Do you get the same answer for the area? What do the angles add up to?
----really stuck on this.

sorry about the length, any help appreciated.

cheers
Scott

Answer
Hi Scott,

Boy, it took me a while to get through this!!  It's been awhile since I've done a lot of vector work, but I think I've got some solutions for you.

On question #1, you are right on the money in parts a and b.  In part c, I think what you want to use is the formula

       | u X v | = |u| * |v| * sin A  (where A is the angle between u and v.   

Since you found | u X v | = 5, you should have

               5 = sqrt(29) * sqrt(6) * sin A

and I got 22.3 degrees for A.


On this one, there's a lot of work to do, and some of it is repetitive.

If we take the points given, we can calculate these vectors:

PR = u = (2-1)i +(1-0)j + (2-1)k = i + j + k

PQ = v = (-1-1)i + (1-0)j + (0-1)k = -2i + j - k

Then, using the dot product formula,

           u * v = |u| |v| cos P

we get       -2 = sqrt(3) sqrt(6) cos P, and I found

               P = 118.1 degrees.

Then to find the area, you can use the trig formula

              Area = 1/2 * side * side * sin(incl angle)

so              A   = 1/2 * |u| |v| sin 118.1

                   = 1/2 * sqrt(3) * sqrt(6) * sin 118.1

                   = 1.87.



For the angles at Q and R, just do the same thing that we did for the angle at P.  I called the vector from Q to R
vector w, and so for angle Q, the vectors forming angle Q are:

w = 3i + 0j + 2k   and -v = 2i -j + k

so v * w = |v| |w| cos Q

  8      =  sqrt(6) sqrt(13) cos Q

and I got Q = 25.1 degrees.  


Similarly, at R,

the vectors forming angle R are -u and -w, so

-u * -w = |u| |w| cos R

   5   =  sqrt(3) sqrt(13) cos R

and  R = 36.8 degrees

which I was pleasantly surprised to find was exactly what it should be if you add the first two and subtract from 180.

If you use the same area formula as I did above, I checked it out using angle Q:

  Area = 1/2 * |v| |w| sin 25.1 = 1.87

so, it does come out as expected.  I didn't bother to do the area using angle R, but I'm sure it is the same.

I sure hope that I did these clearly enough for you to make out what I did along the way.  If not, feel free to write back, and I'll see if I can be more specific.

Hope this helps you out!!

Steve Holleran