Let the angle θ,0<θ<2π between two unit vectors a^ and b^ be sin−1(965). If the vector c=3a^+6b^+9(a^×b^), then the value of 9(c⋅a^)−3(c⋅b^) is
Solutions · 1
Official solution
To solve the problem, we begin by examining the vector c=3a^+6b^+9(a^×b^).
Let's calculate:
c⋅a^=3+6(a^⋅b^).
c⋅b^=3(a^⋅b^)+6.
We need to find the value of 9(c⋅a^)−3(c⋅b^).
Substituting the expressions, we get:
9(c⋅a^)=9(3+6a^⋅b^)=27+54(a^⋅b^)3(c⋅b^)=3(3a^⋅b^+6)=9a^⋅b^+18
Therefore,
9(c⋅a^)−3(c⋅b^)=(27+54(a^⋅b^))−(9a^⋅b^+18)
Simplifying,
=27−18+54(a^⋅b^)−9(a^⋅b^)=9+45(a^⋅b^)
Given sinθ=965, and knowing that for unit vectors cosθ=a^⋅b^, we use the identity (cosθ)2=1−(sinθ)2:
(a^⋅b^)2=1−(965)2=1−8165=8116
Thus, a^⋅b^=94.
Substitute back into the equation:
9+45×94=9+20=29
Therefore, the value is 29.