If ABbar=2abar +bbar and ADbar=abar-2bbar where |abar|=1,|bbar|=1,(abar,bbar)=60degrees are the adjacent sides of a parallelogram,then the length of the diagonal BDbar is

We are given the vectors:

AB⃗ = 2a⃗ + b⃗
AD⃗ = a⃗ - 2b⃗

These represent adjacent sides of a parallelogram. The diagonal BD⃗ can be found using the vector addition property of a parallelogram:

BD⃗ = AB⃗ - AD⃗

Substituting the given values:

BD⃗ = (2a⃗ + b⃗) - (a⃗ - 2b⃗)
= 2a⃗ + b⃗ - a⃗ + 2b⃗
= (2a⃗ - a⃗) + (b⃗ + 2b⃗)
= a⃗ + 3b⃗

Now, to find the magnitude |BD⃗|:

|BD⃗| = |a⃗ + 3b⃗|

Using the magnitude formula for two vectors:

|P⃗ + Q⃗| = √(|P⃗|² + |Q⃗|² + 2|P⃗||Q⃗| cosθ)

Substituting |a⃗| = 1, |b⃗| = 1, and θ = 60°:

|BD⃗| = √(1² + 3² + 2(1)(3)cos60°)
= √(1 + 9 + 6 × (1/2))
= √(1 + 9 + 3)
= √13

Thus, the length of the diagonal BD⃗ is √13.