We study addition and subtraction of vectors and multiplication of a vector by a real number.
Sum of Vectors (Triangle Law)
Connect the initial point of →b to the terminal point of →a. The vector from the initial point to the final terminal point is →a + →b.
→AB + →BC = →AC (triangle law)
- Commutative law: →a + →b = →b + →a
- Associative law: (→a + →b) + →c = →a + (→b + →c)
- →a + →0 = →a, →a + (−→a) = →0
Difference of Vectors
→a − →b = →a + (−→b). Viewed from the same initial point O, →OA − →OB = →BA.
Example ①
Let →OA = →a, →OB = →b, and →OC = →c. Express the following using →a, →b, and →c.
(1) →AB (2) →AC − →BC
Solution:(1) →AB = →OB − →OA = →b − →a
(2) (→c − →a) − (→c − →b) = →b − →a
Let →OA = →a, →OB = →b, and →OC = →c. Express the following using →a, →b, and →c.
(1) →AB (2) →AC − →BC
Solution:(1) →AB = →OB − →OA = →b − →a
(2) (→c − →a) − (→c − →b) = →b − →a
Scalar Multiplication of Vectors
For k→a, if k＞0 it has the same direction as →a and magnitude k|→a|; if k＜0 it has the opposite direction and magnitude |k||→a|; if k=0 it is →0.
- Associative law: k(l→a) = (kl)→a
- Distributive laws: (k + l)→a = k→a + l→a, k(→a + →b) = k→a + k→b
Example ②
Calculate 2(→a + 3→b) − 3(→a − →b).
Solution:2→a + 6→b − 3→a + 3→b = −→a + 9→b
Calculate 2(→a + 3→b) − 3(→a − →b).
Solution:2→a + 6→b − 3→a + 3→b = −→a + 9→b
Condition for Parallel Vectors
When →a ≠ →0 and →b ≠ →0, →a ∥ →b ⇔ there exists a real number k such that →b = k→a.
Key Point
- →AB + →BC = →AC (connect them) / →OA − →OB = →BA
- The direction of k→a is determined by the sign of k, and its magnitude is multiplied by |k|
- The distributive and associative laws work just as they do in numerical calculations
- Parallel condition: there is a real number k such that →b = k→a
Practice Problems
- Simplify 3(2→a − →b) − 2(→a − 3→b).
- When →OA=→a, →OB=→b, and →OC=→c, express →BA + →AC using →a, →b, and →c.
- Express, in terms of →a, a vector parallel to →a with magnitude 5 (two ways).
Solutions and Explanations
- Solution:4→a + 3→b
Explanation: 6→a − 3→b − 2→a + 6→b = 4→a + 3→b. - Solution:→c − →b
Explanation: →BA + →AC = →BC = →c − →b. - Solution:± 5→a/|→a|
Explanation: The unit vector in the direction of →a multiplied by 5; there are two possibilities because of the opposite direction.