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Mathematics Question on Vectors

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

A

yz−yz- plane

B

z−z- axis

C

y−y- axis

D

x−x- axis

Answer

x−x- axis

Explanation

Solution

L→=r→×P→=∣i^j^k^ 12−1 34−2 ∣\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{P}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 2 & -1 \\\ 3 & 4 & -2 \\\ \end{matrix} \right|
L→=i^(−4+4)−j^(−2+3)+k^(4−6)\overrightarrow{L}=\hat{i}(-4+4)-\hat{j}(-2+3)+\hat{k}(4-6)
=−j^−2k^=-\hat{j}-2\hat{k} L→\overrightarrow{L}
has components along -y axis and −z-z axis. The angular momentum is in y-z plane ie, perpendicular to x−x- axis