## anonymous one year ago Find the area of the parallelogram with corner $$P_1$$ and adjacent sides $$P_1 P_2$$ and $$P_1 P_3$$ $$P_1 =(1,2,0)$$, $$P_2= (-1,4,2)$$, $$P_3=(-3,-1,3)$$ If u and v are nonzero vectors in space, then|$$\| u\times v\|$$ is the area of the parallelogram having u and v as adjacent sides. u = $$P_1\space P_2 = -2i+2j+2k$$ How is u equaling that? I thought we times the P1 and p2 but it does not seem to be that way...

1. anonymous

I can figure out the rest but not sure how they got u

2. anonymous

Find the area of the parallelogram with corner $$P_1$$ and adjacent sides$$P1 P2$$ and $$P1 P3$$ $$P_1 =(1,2,0)$$$$P_2= (-1,4,2)$$ $$P_3=(-3,-1,3)$$ If u and v are nonzero vectors in space, then$$\|u\times v\|$$ is the area of the parallelogram having u and v as adjacent side Find the two adjacent sides of this parallelogram . $$u =P_1\space P_2 = -2i+2j+2k$$ $$v =P_1\space P_3 = -4i-3j+3k$$ How are they getting u and V up above?

3. anonymous

21

4. anonymous

@Gunboss is that how old you are? :-D

5. anonymous

@Nixy yep

6. nincompoop

try drawing/graphing it

7. anonymous

Nope that will not help but I think I just thought of something, HAHAHA and it might work. BRB let going to go see

8. anonymous

Nope, that was not it........

9. anonymous

AH I think I got it this time

10. anonymous

BRB

11. anonymous

I GOT IT!!!!! I LOVE IT I LOVE IT $\huge u= (x_2-x_1)+ (y_2-y_1)+(z_2-z_1)$ $\huge u = (i_2-i_1)+ (j_2-j_1)+(k_2-k_1)$ $\huge u = (-1-1)i+ (4-2)j+(2-0)k$ $\huge u = -2i+ 2j+2k$

12. anonymous

I can finish now :-) This stuff is FUN :-D

13. anonymous

After it is all done the $$\| u \times v \| = 2\sqrt{86}$$ Yep got it. Thanks to all that came to look at my problem.