Vector Algebra Question 69
Question: For any vector $ \overset{\to }{\mathop{p}}, $ , the value of $ \frac{3}{2}{ |\overset{\to }{\mathop{p}},\times \hat{i}{{|}^{2}}+|\overset{\to }{\mathop{p}},\times \hat{j}{{|}^{2}}+|\overset{\to }{\mathop{p}},\times \hat{k}{{|}^{2}} } $ is
Options:
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}}, $
Show Answer
Answer:
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}{ |\overset{\to }{\mathop{p}},\times \hat{i}{{|}^{2}}+|\overset{\to }{\mathop{p}},\times \hat{j}{{|}^{2}}+|\overset{\to }{\mathop{p}},\times \hat{k}{{|}^{2}} } $ $ =3(p_1^{2}+p_2^{2}+p_3^{2})=3_p^{\to 2} $
BETA
