Question

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...

Answer 1

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

Answer 2

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?

Answer 3

21

Answer 4

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

Answer 5

@Nixy yep

Answer 6

try drawing/graphing it

Answer 7

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

Answer 8

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

Answer 9

AH I think I got it this time

Answer 10

BRB

Answer 11

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

Answer 12

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

Answer 13

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.