Expert: Paul Klarreich - 6/21/2009

Question
So out of 5 assignments these were the only ones i got stuck on. Can you show me the steps??

1. Find the length of v.
Initial point of v:(0,-1,0)
Terminal point of v;(1,2,-2)

2. Find the angle theta between the two vectors
 u=10i+40j
 v=-3j+8k

3. Find u x v and show that it is orthogonal to both u and v.  u=6k  v=-i+3j+k

4. Find the area of the parallelogram that has vectors as adjacent sides. u=i+3j+k   v=i+k

Answer
Questioner:   Brielle
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  Vectors
Question:  So out of 5 assignments these were the only ones i got stuck on. Can you show me the steps??

1. Find the length of v.
Initial point of v:(0,-1,0)
Terminal point of v;(1,2,-2)

2. Find the angle theta between the two vectors
u=10i+40j
v=-3j+8k

3. Find u x v and show that it is orthogonal to both u and v.  u=6k  v=-i+3j+k

4. Find the area of the parallelogram that has vectors as adjacent sides. u=i+3j+k   v=i+k
..............................................
Hi, Brielle,

1. The length of a vector is sqrt(x^2 + y^2 + z^2), so if v goes from
(0,-1,0) to (1,2,-2), then  v = <1-0,2-(-1),-2-0> = <1,3,-2>, and there's your <x,y,z>

Looks like sqrt(9) = 3.
.......................
2.  u dot v = |u| |v| cos t.

|u| = 10 sqrt(17)
|v| = sqrt(9 + 64) = sqrt(73)

u dot v = -30 + 320 = 290.

290 = 10 sqrt(17)sqrt(73) cos t.
            29
cos t = -----------------
       sqrt(17)sqrt(73)

Now just use your calculator.
..............................
3. u=6k  v=-i+3j+k

I use this layout for  u X v:

| i  j  k  |
| 0  0  6  |
|-1  3  1  |

= -18i - 6j + 0k

Now just dot that with  u and with v.  You should get zero each time.
...............................
4. Isn't the area of the parallelogram equal to | U X V | ?

Use u = i + 3j + k  and    v= i + k,  compute the cross product, then determine its length the way you did it in the first two examples.

| i  j  k  |
| 1  3  1  |
| 1  0  1  |

= <3, 0, -3> =>  3 sqrt(2) length = area.