Try problems that combine what we have learned so far: addition, subtraction, scalar multiplication, components, parallel conditions, and decomposition.
Comprehensive Component Calculations
Example ①
When →a=(2, −1) and →b=(1, 3), find |→a + →b| and the unit vectors parallel to →a − 2→b.
Solution: →a+→b=(3, 2), with magnitude √13.
→a−2→b=(0, −7), with magnitude 7. The unit vectors are (0, ±1)（two possibilities）.
When →a=(2, −1) and →b=(1, 3), find |→a + →b| and the unit vectors parallel to →a − 2→b.
Solution: →a+→b=(3, 2), with magnitude √13.
→a−2→b=(0, −7), with magnitude 7. The unit vectors are (0, ±1)（two possibilities）.
Applications of Parallel Conditions
Example ②
→a=(1, 2), →b=(3, −1). Find t such that →a + t→b is parallel to →a − →b.
Solution: →a+t→b=(1+3t, 2−t), →a−→b=(−2, 3).
Parallel condition: (1+3t)×3 − (2−t)×(−2)=0 → 7t+7=0 → t=−1.
→a=(1, 2), →b=(3, −1). Find t such that →a + t→b is parallel to →a − →b.
Solution: →a+t→b=(1+3t, 2−t), →a−→b=(−2, 3).
Parallel condition: (1+3t)×3 − (2−t)×(−2)=0 → 7t+7=0 → t=−1.
Condition for Three Points to Be Collinear
Three points A, B, C are collinear ⇔ there is a real number k such that →AC = k→AB.
Example ③
Find k such that the three points A(1, 2), B(3, 5), and C(7, k) are collinear.
Solution: →AB=(2, 3), →AC=(6, k−2). Parallel condition: 2(k−2)−3×6=0 → 2k=22 → k=11.
Find k such that the three points A(1, 2), B(3, 5), and C(7, k) are collinear.
Solution: →AB=(2, 3), →AC=(6, k−2). Parallel condition: 2(k−2)−3×6=0 → 2k=22 → k=11.
Key Points
- Calculate components separately; magnitude is √(x²+y²)
- Parallel condition: the cross-product difference of the components is 0
- Three collinear points: →AB ∥ →AC
- Parallel unit vectors: ±→a/|→a|
Practice Problems
- When →a=(−1, 4) and →b=(3, 2), find |2→a − →b|.
- When →a=(4, k) and →b=(−2, 3) are parallel, find k.
- When the three points A(2, 1), B(5, 7), and C(k, 11) are collinear, find k.
Answers and Explanations
- Answer: √61
Explanation: 2→a−→b=(−5, 6), with magnitude √(25+36)=√61. - Answer: k=−6
Explanation: 4×3 − k×(−2)=0 → 12+2k=0 → k=−6. - Answer: k=7
Explanation: →AB=(3, 6), →AC=(k−2, 10). 3×10−6(k−2)=0 → 42=6k → k=7.