Cho hai đường thẳng denta: (x - 1)/2 = (y + 3)/1 = (z - 4)/-2 denta': (x + 2)/4 = (y - 1)/-2 = (z + 1)/4
Giải thích
Ta có M0M'0→ = (−3; 4; −5)
a→ = (2; 1; −2)
n→ = M0M'0→ ∧ a→ = (−3; −16; −11)
Ta có M0M'0→ = (−3; 4; −5)
a→ = (2; 1; −2)
n→ = M0M'0→ ∧ a→ = (−3; −16; −11)