(a + b)^2 = (c)^2 ⇒ a^2 + b^2 + 2a·b = c^2 ⇒ 3^2 + 4^2 + 2·4·3 = 5^2 ⇒ 2a·b = 0 or a·b = 0. a ⊥ b. Here a^2 + b^2 = c^2. Hence, a ⊥ b.

(a) (a + b)^2 = (c)^2
(b) Given: x = 5t − 2t^2, y = 10t, vx = dx/dt = 5 − 4t, vy = dy/dt = 10, ax = dvx/dt = −4, ay = dvy/dt = 0, a = axi + ayj, a = −4m/s^2. Hence, acceleration of particle at (t = 2s) = −4m/s^2
(c) Let θ be the angle which the particle makes with x-axis. From figure, tan θ = 3/√3 = √3 ⇒ θ = tan^−1(√3) = 60°
(d) If a vector r in X–Y plane then its orthogonal vector Rx = R cos θ and Ry = R sin θ. R sin θ = tan θ = Ry/Rx

Correct answer: (a) (a + b)^2 = (c)^2

Solution

The given derivation in option (a) is incorrect because (a + b)^2 = c^2 does not imply a^2 + b^2 + 2ab = c^2 unless a and b are orthogonal. The conclusion a ⊥ b is invalid.

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