Kinematics Question 293

Question: The position of a particle is given by $ \overrightarrow{r}=(\overrightarrow{i}+2\overrightarrow{j}-\overrightarrow{k}) $ momentum $ \overrightarrow{P}=(3\overrightarrow{i}+4\overrightarrow{j}-2\overrightarrow{k}). $ The angular momentum is perpendicular to [EAMCET (Engg.) 1998]

Options:

A) $ \sqrt{61} $ sq. unit

B) $ \sqrt{59} $ sq. unit

C) $ \sqrt{49} $ sq. unit

D) $ \sqrt{52} $ sq. unit

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Answer:

Correct Answer: B

Solution:

$ \vec{A}=\hat{j}+3\hat{k} $ , $ \vec{B}=\hat{i}+2\hat{j}-\hat{k} $

$ \vec{C}=\vec{A}\times \vec{B} $

$ = \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 0 & 1 & 3 \\ 1 & 2 & -1 \\ \end{vmatrix} $

$ =-7\hat{i}+3\hat{j}-\hat{k} $

Hence area = $ |\vec{C}|=\sqrt{49+9+1}=\sqrt{59}squnit $



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