Kinematics Question 293
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}\ \text{sq unit} $
BETA
