A) 1
B) \[\pm \,4\]
C) 3
D) ? 2
Correct Answer: B
Solution :
If angle between \[\mathbf{b}\] and \[\mathbf{c}\] is \[\alpha \] and \[|\mathbf{b}\times \mathbf{c}|=\sqrt{15}\] \[|\mathbf{b}|\] \[|\mathbf{c}|\] \[\sin \alpha \] = \[\sqrt{15}\] \[\sin \alpha =\frac{\sqrt{15}}{4}\] ; \[\therefore \] \[\cos \alpha =\frac{1}{4}\] \[\mathbf{b}-2\mathbf{c}\] = \[\lambda \,\mathbf{a}\Rightarrow |\mathbf{b}-2\mathbf{c}{{|}^{2}}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[|\mathbf{b}{{|}^{2}}+4|\mathbf{c}{{|}^{2}}-4\,.\,\mathbf{b}\,.\,\mathbf{c}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[16+4-4\{|\mathbf{b}|\,|\mathbf{c}|\cos \alpha \}={{\lambda }^{2}}\] \[16+4-4\times 4\times 1\times \frac{1}{4}\]\[={{\lambda }^{2}}\Rightarrow {{\lambda }^{2}}=16\Rightarrow \lambda =\pm 4\].
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