3 ms·

(The second definition is also wrong – it's true only for unit vectors.)

• jimbob45 5y ago
  If you express your vectors as scalar multiples of unit vectors, it works, right?
  • Jensson 5y ago
    Then you'd still multiply those scalars. In your example you just replace 1 and -1 with the product of the vectors length.
• dorkwood 5y ago
  That's why I wrote "as long as they're unit vectors" in my definition.
  • wizzwizz4 5y ago
    I swear I double-checked the paragraph before replying… Sorry for the “correction”.
    • dorkwood 5y ago
      No worries! I probably shouldn't have put it in parentheses at the end...