The position vector of 1 kg object is r→=(3i^−j^)m and its velocity v→=(3j^+k^)ms−1. The magnitude of its angular momentum is xNm where x is

Using L→=r→×p→=r→×mv→,m=1kgL→=(3i^−j^)×(3j^+k^)=(9k^−3j^−i^)N−s⇒|L¯|=91N−s

The position vector of 1 kg object is r→=(3i^−j^)m and its velocity v→=(3j^+k^)ms−1. The magnitude of its angular momentum is xNm where x is

Using L→=r→×p→=r→×mv→,m=1kgL→=(3i^−j^)×(3j^+k^)=(9k^−3j^−i^)N−s⇒|L¯|=91N−s

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The position vector of 1 kg object is $\vec{r} = (3 \hat{i} - \hat{j}) m$ and its velocity $\vec{v} = (3 \hat{j} + \hat{k}) {ms}^{- 1}$ . The magnitude of its angular momentum is $\sqrt{x} Nm$ where x is

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Detailed Solution

$\begin{array}{l} Using \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m \vec{v}, m = 1 kg \\ \vec{L} = (3 \hat{i} - \hat{j}) \times (3 \hat{j} + \hat{k}) = (9 \hat{k} - 3 \hat{j} - \hat{i}) N - s \\ \Rightarrow | \bar{L} | = \sqrt{91} N - s \end{array}$