Expert: Paul Klarreich - 4/21/2007

Question
Hi there im stuck on these questions and i was wondering if you could show me how they are done. I've missed a few classes due to sickness and trying random questions out of a textbook with no answers or working. 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

(b) the area of the parallelogram defined by u and v

(c) the sine of the angle between u and v


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.


Answer
Questioner:   Scott
Category:  Advanced Math
 
Subject:  algebra questions.
Question:  Hi there im stuck on these questions and i was wondering if you could show me how they

are done. I've missed a few classes due to sickness and trying random questions out of a textbook with no answers or working. 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

(b) the area of the parallelogram defined by u and v

(c) the sine of the angle between u and v


2. Consider the triangle in 3-space which has as vertices the points P,Q,R with
respective position vectors 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.
.....................................................
Hi, Scott,

It has been a while, but I recall some aspects of the vector (a.k.a. cross) product of two vectors:

If  A and B are two vectors in space, then:

(Notation:  | A | is the length of vector A.)

1. A X B is a vector that is orthogonal (perpendicular) to both of A and B.

2. If @ is the angle between vectors A and B, then:

   | A X B | = | A || B | sin @

2a. | A X B | is the area of the parallelogram determined by vectors A and B.
. . . . . . . . . . . . . . . . .

We can handle 1 now:

1. let u=2i-3j+4k, v=i-j+2k. Find(using the cross product):
(a) a vector perpendicular to both u & v

By rule 1, that will be their cross product.  I like this notation for the cross product:
Determinant:
|  i   j   k  |
|  2  -3   4  |
|  1  -1   2  |

= (-6 + 4)i + (4 - 4)j + (-2 + 3)k
= -2i + 0j + k

Check:  (-2i + 0j + k) dot (2i - 3j + 4k) = -4 + 0 + 4 = 0
and:    (-2i + 0j + k) dot (1i - 1j + 2k) = -2 + 0 + 2 = 0

(b) the area of the parallelogram defined by u and v

That's the length of the cross-product:

sqrt(4 + 0 + 1) = sqrt(5)

(c) the sine of the angle between u and v

Use rule 2:| u X v | = | u || v | sin @

| u | = sqrt(4 + 9 + 16) = sqrt(29)
| v | = sqrt(1 + 1 + 4)  = sqrt(6)

sqrt(5) = sqrt(29) sqrt(6) sin @
           sqrt(5)
sin @ = ----------------
       sqrt(29) sqrt(6)
...........................................
  
Example 2:

P(1,0,1) ,
Q(-1,1,0),
R(2,1,2)

are your points.

Then vectors:
PQ = [-2, 1, -1] and
PR = [1, 1, 1]
define your triangle.

Using that same stuff:

| PQ | = sqrt(4 + 1 + 1) = sqrt(6)
| PR | = sqrt(1 + 1 + 1) = sqrt(3)

Now the cross product is:

|  i   j   k  |
| -2   1  -1  |
|  1   1   1  |

= (1 + 1)j + (-1 + 2)j + (-2 -1)k
= 2i + j - 3k

Check: (2,1,-3)(-2,1,-1) = -4 + 1 + 3 = 0
Check: (2,1,-3)(1,1,1) = 2 + 1 - 3 = 0

OK, we are in business.

| PQ X PR | = sqrt(4 + 1 + 9) = sqrt(14)

           sqrt(14)
sin P = ----------------
       sqrt(6) sqrt(3)

       sqrt(14)
sin P = --------
       sqrt(18)

       sqrt(7)
sin P = --------
       sqrt(9)

       sqrt(7)
sin P = --------
          3

Now use your calculator:

sin P = 0.8819

and  P = 61.87 degrees.

I think now you can compute the area of the triangle (one-half the parallelogram) and use this same method to work out angles Q and R.