A) \[\pm \frac{1}{4}\]
B) \[\pm \frac{1}{2}\]
C) \[\pm \,\,1\]
D) None of the above
Correct Answer: C
Solution :
\[\mathbf{a}\times \mathbf{b}=2\mathbf{a}\times \mathbf{c}\] \[\Rightarrow \] \[\mathbf{a}\times \mathbf{b}-2\mathbf{a}\times \mathbf{c}=0\] \[\Rightarrow \] \[\mathbf{a}\times (\mathbf{b}-2\mathbf{c})=0\] \[\Rightarrow \] \[\mathbf{a}\]and\[(\mathbf{b}-2\mathbf{c})\]are collinear. \[\Rightarrow \] \[\mathbf{b}-2\mathbf{c}=\lambda \mathbf{a}\] \[\Rightarrow \] \[|\mathbf{b}-2\mathbf{c}{{|}^{2}}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[\Rightarrow \] \[|\mathbf{b}{{|}^{2}}+4|\mathbf{c}{{|}^{2}}-4\,\,\mathbf{b}\cdot \mathbf{c}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[\Rightarrow \] \[16+4-4(4)(1)\frac{1}{4}={{\lambda }^{2}}(1)\] \[\therefore \] \[\lambda =\pm 4\]You need to login to perform this action.
You will be redirected in
3 sec