Janek191:
→
u = 3 p − q = 3*[1,0] − [ 0, 1] = [ 3 , 0 ] − [ 0, 1] = [ 3, − 1]
→
v = 5 p+ 2 q = 5*[ 1, 0] + 2 *[ 0, 1] = [ 5, 0 ] + [ 0 , 2] = [ 5, − 2]
u =
√ 32 + (−1)2 =
√10
v =
√ 52 + (−2)2 =
√29
→ →
u o v = 3*5 − 1*( −2) = 15 + 2 = 17
więc
| | 17 | | 17 | |
cos α = |
| = |
| |
| | √10*√29 | | √290 | |
więc
zatem pole równoległoboku
| | 1 | |
P = u*v*sin α = √10*√29 * |
| = 1 |
| | √290 | |
====================================
u*h
1 = P
oraz
v*h
2 = P
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