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Physics Question on System of Particles & Rotational Motion

A particle has the position vector r=i^−2j^+k^{r} = \hat{i} - 2\hat{j } + \hat{k} and the linear momentum p=2i^−j^+k^{p} = 2\hat{i} - \hat{j} + \hat{k}. Its angular momentum about the origin is

A

−i^+j^−3k^- \hat{i} + \hat{j} - 3 \hat{k}

B

−i^+j^+3k^- \hat{i} + \hat{j} + 3 \hat{k}

C

i^−j^+3k^\hat{i} - \hat{j} + 3 \hat{k}

D

i^−j^−5k^\hat{i} - \hat{j} - 5 \hat{k}

Answer

−i^+j^+3k^- \hat{i} + \hat{j} + 3 \hat{k}

Explanation

Solution

Given, r=i^−2j^+k^r =\hat{ i }-2 \hat{ j }+\hat{ k } p=2i^−j^+k^p =2 \hat{ i }-\hat{ j }+\hat{ k } Angular momentum J=r×pJ = r \times p =(i^−2j^+k^)×(2i^−j^+k^)=(\hat{ i }-2 \hat{ j }+\hat{ k }) \times(2 \hat{ i }-\hat{ j }+\hat{ k }) J=∣i^j^k^ 1−21 2−11∣J =\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\\ 1 & -2 & 1 \\\ 2 & -1 & 1\end{vmatrix} J=i^(−2+1)−j^(1−2)+k^(−1+4)J =\hat{ i }(-2+1)-\hat{ j }(1-2)+\hat{ k }(-1+4) J=−i^+j^+3k^J =-\hat{ i }+\hat{ j }+3 \hat{ k }