A force F→=ai^+3j^+6k^ is acting on a particle having position vector r→=2i^−6j^−12k^ w.r.t origin. The value of “a” for which the angular momentum of the particle about origin is conserved, is

τ=0 τ=r¯×p¯r¯×p¯=0ijk2−6−12a36=0i(−36+36)−j(12+12a)+k(6+6a)=0i+0j+0ki(0)−j(12+12a)+k(6+6a)=0i+0j+0k12+12a=012a=−12a=−1212=−1

A force F→=ai^+3j^+6k^ is acting on a particle having position vector r→=2i^−6j^−12k^ w.r.t origin. The value of “a” for which the angular momentum of the particle about origin is conserved, is

τ=0 τ=r¯×p¯r¯×p¯=0ijk2−6−12a36=0i(−36+36)−j(12+12a)+k(6+6a)=0i+0j+0ki(0)−j(12+12a)+k(6+6a)=0i+0j+0k12+12a=012a=−12a=−1212=−1

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A force $\vec{F} = a \hat{i} + 3 \hat{j} + 6 \hat{k}$ is acting on a particle having position vector $\vec{r} = 2 \hat{i} - 6 \hat{j} - 12 \hat{k}$ w.r.t origin. The value of “a” for which the angular momentum of the particle about origin is conserved, is

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Detailed Solution

$\begin{array}{l} τ = 0 τ = \bar{r} \times \bar{p} \\ \bar{r} \times \bar{p} = 0 \\ |\begin{array}{ccc} i & j & k \\ 2 & - 6 & - 12 \\ a & 3 & 6 \end{array}| = 0 \\ i (- 36 + 36) - j (12 + 12 a) + k (6 + 6 a) = 0 i + 0 j + 0 k \\ i (0) - j (12 + 12 a) + k (6 + 6 a) = 0 i + 0 j + 0 k \\ 12 + 12 a = 0 \\ 12 a = - 12 \\ a = \frac{- 12}{12} = - 1 \end{array}$