anonymous
  • anonymous
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...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I can figure out the rest but not sure how they got u
anonymous
  • 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?
anonymous
  • anonymous
21

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anonymous
  • anonymous
@Gunboss is that how old you are? :-D
anonymous
  • anonymous
@Nixy yep
nincompoop
  • nincompoop
try drawing/graphing it
anonymous
  • 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
anonymous
  • anonymous
Nope, that was not it........
anonymous
  • anonymous
AH I think I got it this time
anonymous
  • anonymous
BRB
anonymous
  • anonymous
I GOT IT!!!!! I LOVE IT I LOVE IT \[\huge u= (x_2-x_1)+ (y_2-y_1)+(z_2-z_1) \] \[ \huge u = (i_2-i_1)+ (j_2-j_1)+(k_2-k_1) \] \[ \huge u = (-1-1)i+ (4-2)j+(2-0)k \] \[ \huge u = -2i+ 2j+2k \]
anonymous
  • anonymous
I can finish now :-) This stuff is FUN :-D
anonymous
  • 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.

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