Theory question 114
Vectors(a) Given that x = 3i − j, y = 2i + kj and the cosine of the angle between x and y is √5/5, find the values of the constant k.
(b) In the quadrilateral ABCD, AB = (5, −1), AC = (−6, −9) and BD = (4, −7) (column vectors). Show whether or not ABCD is a parallelogram.
Show the model answer
Model answer
(a) x·y = 6 − k; |x| = √10, |y| = √(4 + k²).
(6 − k)/(√10√(4 + k²)) = 1/√5 gives (6 − k)² = 2(4 + k²), k² + 12k − 28 = 0.
k = 2 or k = −14.
(b) Take A as origin: B = (5, −1), C = (−6, −9), D = B + BD = (9, −8).
DC = C − D = (−15, −1) ≠ AB = (5, −1) (also AD = (9, −8) ≠ BC = (−11, −8)).
Opposite sides are not equal vectors, so ABCD is not a parallelogram.
Explanation
A parallelogram ABCD needs AB = DC.