Mila:
l:
x+y−z=0
x−y+z=2
przyjmujesz z jako parametr, z=t, t∊R
x+y=t
x−y=2−t
−−−−−−−
2x=2
x=1
1+y=t⇔y=−1+t
l:
x=1
y=−1+t
z=t, A=(1,−1,0)∊l i k
→=[0,1,1] wektor kierunkowy prostej l,
P=(3,4,5)
PA
→=[−3,−3,−4]
| |PA→ x k→ || | |
d(P,l)= |
| = |
| |k| | |
| [−3,−3,−4] x [0,1,1]| | |
= |
| = |
| √02+12+12 | |
| |[1,3,−3]| | | √12+32+32 | | √19 | | √38 | |
= |
| = |
| = |
| = |
| |
| √2 | | √2 | | √2 | | 2 | |