Let a⃗ =2i^+2j^+k^𝑎→=2𝑖^+2𝑗^+𝑘^ and b⃗ 𝑏→ be another vector such that a⃗ .b⃗ =14𝑎→.𝑏→=14 and a⃗ ×b⃗ =3i^+j^−8k^𝑎→×𝑏→=3𝑖^+𝑗^−8𝑘^ the vector b⃗ 𝑏→ =

The vector b⃗ can be found using the formula for the dot product and the cross product.

The dot product of two vectors a⃗ and b⃗ is given by a⃗ .b⃗ = |a⃗||b⃗|cosθ, where |a⃗| and |b⃗| are the magnitudes of a⃗ and b⃗ respectively, and θ is the angle between them.

Given that a⃗ .b⃗ = 14, we can find |b⃗|cosθ = 14/|a⃗|.

The magnitude of a⃗ is √(2^2 + 2^2 + 1^2) = √9 = 3.

So, |b⃗|cosθ = 14/3.

The cross product of two vectors a⃗ and b⃗ is given by a⃗ ×b⃗ = |a⃗||b⃗|sinθn^, where n^ is the unit vector perpendicular to both a⃗ and b⃗.

Given that a⃗ ×b⃗ = 3i^+j^−8k^, we can find |b⃗|sinθ = √((3)^2 + (1)^2 + (-8)^2)/|a⃗| = √74/3.

Now, we know that |b⃗| = √((|b⃗|cosθ)^2 + (|b⃗|sinθ)^2) = √((14/3)^2 + (74/3)^2) = √(196+74)/3 = √90.

So, the vector b⃗ can be represented as b⃗ = |b⃗|(cosθi^ + sinθj^ + 0k^), where cosθ and sinθ can be found from the dot product and cross product respectively.

Please note that this is a general solution and the actual vector b⃗ will depend on the specific values of θ.