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Question: The position vectors of radius are \(2\widehat{i} + \widehat{j} + \widehat{k}\) and \(2\widehat{i} -...

The position vectors of radius are 2i^+j^+k^2\widehat{i} + \widehat{j} + \widehat{k} and 2i^−3j^+k^2\widehat{i} - 3\widehat{j} + \widehat{k} while those of linear momentum are 2i^+3j^−k^.2\widehat{i} + 3\widehat{j} - \widehat{k}. Then the angular momentum is :

A

2i^−4k^2\widehat{i} - 4\widehat{k}

B

4i^−8k^4\widehat{i} - 8\widehat{k}

C

2i^−4j^+2k^2\widehat{i} - 4\widehat{j} + 2\widehat{k}

D

4i^−8k^4\widehat{i} - 8\widehat{k}

Answer

4i^−8k^4\widehat{i} - 8\widehat{k}

Explanation

Solution

Radius vector

r→=r2→−r1→=(2i^−3j^+k^)−(2i^+j^+k^)\overrightarrow{r} = \overset{\rightarrow}{r_{2}} - \overset{\rightarrow}{r_{1}} = (2\widehat{i} - 3\widehat{j} + \widehat{k}) - (2\widehat{i} + \widehat{j} + \widehat{k})

∴ r→=−4j^\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{-}\mathbf{4}\widehat{\mathbf{j}}

Linear momentum p→=2i^+3j^−k^\overset{\rightarrow}{p} = 2\widehat{i} + 3\widehat{j} - \widehat{k}

L→=r→×p→=(−4j^)×(2i^+3j^−k^)\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p} = ( - 4\widehat{j}) \times (2\widehat{i} + 3\widehat{j} - \widehat{k})

\widehat{i} & \widehat{j} & \widehat{k} \\ 0 & - 4 & 0 \\ 2 & 3 & - 1 \end{matrix} \right| = 4\widehat{i} - 8\widehat{k}$$