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Question: The position of a particle is given by \(\overrightarrow{r} = \widehat{i} + 2\widehat{j} - \widehat{...

The position of a particle is given by r→=i^+2j^−k^\overrightarrow{r} = \widehat{i} + 2\widehat{j} - \widehat{k} and its linear momentum is given by . p→=3i^+4j^−2k^\overrightarrow{p} = 3\widehat{i} + 4\widehat{j} - 2\widehat{k} Then its angular momentum about the origin is perpendicular to

A

x-axis

B

y-axis

C

z-axis

D

yzplane

Answer

x-axis

Explanation

Solution

Here, r→=i^+2j^−k^,p→=3i^+4j^−2k^\overset{\rightarrow}{r} = \widehat{i} + 2\widehat{j} - \widehat{k},\overrightarrow{p} = 3\widehat{i} + 4\widehat{j} - 2\widehat{k}

\widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 2 & - 1 \\ 3 & 4 & - 2 \end{matrix} \right|$$ $$= \widehat{i}( - 4 + 4) + \widehat{j}( - 3 + 2) + \widehat{k}(4 - 6) = 0\widehat{i} - 1\widehat{j} - 2\widehat{k}$$ $\overset{\rightarrow}{L.}$has components along – y axis and –z axis but it has no component along in the x-axis. The angular momentum $\overset{\rightarrow}{L.}$ is in yz-plane. i.e., perpendicular to x – axis.