A) -4
B) -2
C) 1
D) 3
Correct Answer: A
Solution :
Given that,\[|\overrightarrow{a}|=|\overrightarrow{c}|=1,|\overrightarrow{b}|=4.\] Let angle between \[\overrightarrow{b}\] and \[\overrightarrow{c}\] is\[\alpha \], then \[|\overrightarrow{b}\times \overrightarrow{c}|=\sqrt{15}\] (given) \[\Rightarrow \]\[|\overrightarrow{b}|\,\,|\overrightarrow{c}|\sin \alpha =\sqrt{15}\] \[\Rightarrow \]\[\sin \alpha =\frac{\sqrt{15}}{4\times 1}=\frac{\sqrt{15}}{4}\] \[\therefore \]\[\cos \alpha =\sqrt{1-{{\sin }^{2}}\alpha }=\frac{1}{4}\] Also given, \[\overrightarrow{b}-2\overrightarrow{c}=\lambda \overrightarrow{a}\] On squaring both sides, we get \[{{\overrightarrow{b}}^{2}}+4{{\overrightarrow{c}}^{2}}-4\overrightarrow{b}.\overrightarrow{c}={{\lambda }^{2}}{{\overrightarrow{a}}^{2}}\] \[\Rightarrow \]\[16+4-4|\overrightarrow{b}|\,\,|\overrightarrow{c}|\cos \alpha ={{\lambda }^{2}}\] \[\Rightarrow \]\[16+4-4\times 4\times 1\times \frac{1}{4}={{\lambda }^{2}}\] \[\Rightarrow \]\[{{\lambda }^{2}}=16+4-4=16\]\[\Rightarrow \]\[\lambda =\pm 4\]
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