Given the vector a = (k, 1), B = (6, - 2), if a is parallel to B, then the real number k = 1___ .

Given the vector a = (k, 1), B = (6, - 2), if a is parallel to B, then the real number k = 1___ .


∵ a = (k, 1), B = (6, - 2), and a is parallel to B, ∵ - 2k-6 × 1 = 0, the solution is k = - 3, so the answer is: - 3



Given vector a = (- 1, 2), B = (5, K), if a ‖ B, then the value of real number k is ()
A. 5B. -5C. 10D. -10


∵ a = (- 1, 2), B = (5, K), a ∥ B, ∥ - 1 × K-2 × 5 = 0, the solution is k = - 10