Explanation
Given that the magnitude of each vector is 1 (∣v∣=1), we use the dot product property ∣v∣2=v⋅v=1.
1. Determine individual dot products:
From the previous analysis:
2. Use the magnitude of the sum a+b+c:
We are given that ∣a+b+c∣=1. Squaring both sides:
∣a+b+c∣2=(a+b+c)⋅(a+b+c)=1
Expanding the dot product:
∣a∣2+∣b∣2+∣c∣2+2(a⋅b+b⋅c+c⋅a)=1
Substitute the known values (∣a∣=∣b∣=∣c∣=1 and the dot products calculated above):
1+1+1+2(−21−21+c⋅a)=1
3+2(−1+c⋅a)=1
3−2+2(c⋅a)=1
1+2(c⋅a)=1
2(c⋅a)=0⟹c⋅a=0
3. Find the angle:
The dot product is defined as a⋅c=∣a∣∣c∣cosθ=0.
Since ∣a∣=1 and ∣c∣=1:
1⋅1⋅cosθ=0
cosθ=0⟹θ=2π
Correct Option: (c)
