Given two unit vectors a and b, we know that: |a|=|b|=1 Consider the vectors c=a+2b and d=5a−4b. These vectors are perpendicular, meaning: c⋅d=0 Let's compute the dot product: (a+2b)⋅(5a−4b)=a⋅(5a−4b)+2b⋅(5a−4b) Expanding this, we have: =a⋅5a−a⋅4b+2b⋅5a−2b⋅4b This simplifies to: =5|a|2−4a⋅b+10a⋅b−8|b|2 Simplifying further by substituting |a|=1 and |b|=1 : =5(1)+6a⋅b−8(1)=0 Which simplifies to: ‌5−8+6(a⋅b)=0 ‌−3+6(a⋅b)=0 Solving for a⋅b : ‌6(a⋅b)=3 ‌a⋅b=‌
3
6
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2
The dot product a⋅b=|a||b|cos‌θ, so: cos‌θ=‌