好緊急:問一問

Let θ be the angle between vector a and vector b.

θ = 60°


a‧b = │a│‧│b│cosθ

a‧b = (4)(5)cos60°

a‧b = 10


Since ka - b and a + 2b are perpendicular.

So, (ka - b)‧(a + 2b) = 0

k│a│² + (2k - 1)a‧b - 2│b│² = 0

k(4)² + (2k - 1)(10) - 2(5)² = 0

16k + 20k - 10 - 50 = 0

k = 5/3

a•b=│a││b│cos fita=4 (5)(cos 60)=10

because 垂直

(ka-b)(a+2b)=0
k│a│^2+2ka•b-a•b-2│b│^2=0
16k+20k-10-50=0
k=60/36=5/3//