Question

two particle emitting from a source of have a displacement r1=4i+3j+8k and r2=2i+10j+5k ,at any time ,calculate to the displacement of the second particle relatives to the first particle?

Answer 1

Displacement of the 2nd particle relative to the first one is -2i + 7j - 3k and its magnitude is 7.875 unit.

Two particles emitting from a source of have a displacement r₁ = (4i + 3j + 8k) and r₂ = (2i + 10j + 5k) , at any time.

We have to find the displacement of the second particle relative to the first particle.

Displacement of 2nd relative to the first is given by, r₂₁ = r₂ - r₁

= (2i + 10j + 5k) - (4i + 3j + 8k)

= (2 - 4)i + (10 - 3)j + (5 - 8)k

= -2i + 7j - 3k

magnitude of relative displacement of 2nd particle is

|r₂₁| = √{(-2)² + (7)² + (3)²}

= √(4 + 49 + 9)

= √62 ≈ 7.875 unit.

Therefore displacement of the 2nd particle relative to the first one is -2i + 7j - 3k and its magnitude is 7.875 unit.