Vector Algebra Question 154
Question: Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle q to both a and b. If $ \mathbf{c}=\alpha ,\mathbf{a}+\beta ,\mathbf{b}+\gamma ,(\mathbf{a}\times \mathbf{b}), $ then
[Orissa JEE 2003]
Options:
A) $ \alpha =\beta =\cos \theta ,{{\gamma }^{2}}=\cos 2\theta $
B) $ \alpha =\beta =\cos \theta ,{{\gamma }^{2}}=-\cos 2\theta $
C) $ \alpha =\cos \theta ,\beta =\sin \theta ,{{\gamma }^{2}}=\cos 2\theta $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \mathbf{c}=\alpha ,\mathbf{a}+\beta ,\mathbf{b}+\gamma ,(\mathbf{a}\times \mathbf{b}) $
$ \Rightarrow \mathbf{c},.,\mathbf{a}=\alpha $ and $ \mathbf{c},.,\mathbf{b}=\beta $
$ \Rightarrow \alpha =\beta =\cos \theta $
Also, $ 1=\mathbf{c},.,\mathbf{c} $ ,
$ \therefore [ \alpha ,\mathbf{a}+\beta ,\mathbf{b}+\gamma ,(\mathbf{a}\times \mathbf{b}) ],.,[ (\alpha ,\mathbf{a}+\beta ,\mathbf{b})+\gamma ,(\mathbf{a}\times \mathbf{b}) ]=1 $
$ \Rightarrow 2{{\alpha }^{2}}+{{\gamma }^{2}}{{(\mathbf{a}\times \mathbf{b})}^{2}}=1 $ ,
$ { \because \alpha ,=,\beta } $
$ \Rightarrow 2{{\alpha }^{2}}+{{\gamma }^{2}}[{{\mathbf{a}}^{2}}{{\mathbf{b}}^{2}}-{{(\mathbf{a}.\mathbf{b})}^{2}}]=1\Rightarrow 2{{\alpha }^{2}}+{{\gamma }^{2}}=1 $
Hence, $ {{\gamma }^{2}}=1-2{{\alpha }^{2}}=1-2,{{\cos }^{2}}\theta =-\cos 2\theta . $
BETA
