A) \[\overset{\to 2}{\mathop{p}}\,\]
B) \[2\overset{\to 2}{\mathop{p}}\,\]
C) \[3\overset{\to 2}{\mathop{p}}\,\]
D) \[4\overset{\to 2}{\mathop{p}}\,\]
Correct Answer: C
Solution :
[c] Suppose \[\overset{\to }{\mathop{p}}\,={{p}_{1}}\hat{i}+{{p}_{2}}\hat{j}+{{p}_{3}}\hat{k}\] \[\overset{\to }{\mathop{p}}\,\,\times \hat{i}={{p}_{2}}\hat{j}\times \hat{i}+{{p}_{3}}\hat{k}\times \hat{i}=-{{p}_{2}}\hat{k}+{{p}_{3}}\hat{j}\] \[|\overset{\to }{\mathop{p}}\,\times \hat{i}{{|}^{2}}=p_{2}^{2}+p_{3}^{2}\] Similarly, \[|\overset{\to }{\mathop{p}}\,\times \hat{j}{{|}^{2}}=p_{3}^{2}+p_{1}^{2},\] \[|\overset{\to }{\mathop{p}}\,\times \hat{k}{{|}^{2}}=p_{1}^{2}+p_{2}^{2},\] \[\therefore \frac{3}{2}\left\{ |\overset{\to }{\mathop{p}}\,\times \hat{i}{{|}^{2}}+|\overset{\to }{\mathop{p}}\,\times \hat{j}{{|}^{2}}+|\overset{\to }{\mathop{p}}\,\times \hat{k}{{|}^{2}} \right\}\] \[=3(p_{1}^{2}+p_{2}^{2}+p_{3}^{2})=3_{p}^{\to 2}\]
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