If 𝑨⃗⃗ = 𝟐𝒊̂ + 𝟑𝒋̂ and 𝑩⃗⃗ = 𝒊̂ − 𝒋̂ then find (i) 𝑨⃗⃗ + 𝑩⃗⃗ (ii) 𝑨⃗⃗ − 𝑩⃗⃗ (iii) 𝑨⃗⃗ . 𝑩⃗⃗⃗ (iv) 𝑨⃗⃗ x 𝑩

(i) To find 𝑨⃗⃗ + 𝑩⃗⃗, we add the corresponding components of vectors 𝑨⃗⃗ and 𝑩⃗⃗.

So, 𝑨⃗⃗ + 𝑩⃗⃗ = (2𝒊̂ + 𝒊̂) + (3𝒋̂ - 𝒋̂) = 3𝒊̂ + 2𝒋̂

(ii) To find 𝑨⃗⃗ - 𝑩⃗⃗, we subtract the corresponding components of vectors 𝑨⃗⃗ and 𝑩⃗⃗.

So, 𝑨⃗⃗ - 𝑩⃗⃗ = (2𝒊̂ - 𝒊̂) + (3𝒋̂ - (-𝒋̂)) = 𝒊̂ + 4𝒋̂

(iii) To find 𝑨⃗⃗ . 𝑩⃗⃗, we multiply the corresponding components of vectors 𝑨⃗⃗ and 𝑩⃗⃗ and then add them.

So, 𝑨⃗⃗ . 𝑩⃗⃗ = (2𝒊̂ * 𝒊̂) + (3𝒋̂ * -𝒋̂) = 2 - 3 = -1

(iv) To find 𝑨⃗⃗ x 𝑩⃗⃗, we use the formula 𝑨⃗⃗ x 𝑩⃗⃗ = |𝑨⃗⃗| |𝑩⃗⃗| sin(θ) 𝑛̂, where |𝑨⃗⃗| and |𝑩⃗⃗| are the magnitudes of 𝑨⃗⃗ and 𝑩⃗⃗, θ is the angle between 𝑨⃗⃗ and 𝑩⃗⃗, and 𝑛̂ is the unit vector perpendicular to both 𝑨⃗⃗ and 𝑩⃗⃗. However, in this case, since 𝑨⃗⃗ and 𝑩⃗⃗ are in the plane, their cross product is a vector perpendicular to the plane, and its components can be found using the determinant of a matrix formed by the components of 𝑨⃗⃗ and 𝑩⃗⃗.

So, 𝑨⃗⃗ x 𝑩⃗⃗ = (2𝒊̂ x 𝒊̂ - 3𝒋̂ x -𝒋̂) = 0𝒌̂ - (-3)𝒌̂ = 3𝒌̂.