A) \[~\pm \,4\]
B) 14
C) \[\pm \text{ }2~\]
D) 12
Correct Answer: A
Solution :
\[a\times b=2(a\times c)\] \[\Rightarrow \]\[a\times (b-2c)=0\Rightarrow a\]a is parallel to \[b-2c.\] Now, \[(b-2c)=\lambda a\Rightarrow |b-2c{{|}^{2}}={{\lambda }^{2}}|a{{|}^{2}}\] \[\Rightarrow \]\[|b{{|}^{2}}+\,4|c{{|}^{2}}-4(b.c)={{\lambda }^{2}}|a{{|}^{2}}\] \[\Rightarrow \]\[16+4-4\times |b||c|\times \frac{1}{4}={{\lambda }^{2}}\] \[\Rightarrow \]\[20-4={{\lambda }^{2}}\Rightarrow \lambda =\pm 4\]You need to login to perform this action.
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