A) -1
B) 1
C) -4
D) 2
Correct Answer: C
Solution :
Let angle between\[\mathbf{b}\]and\[\mathbf{c}\]is\[\alpha \], then \[|\mathbf{b}\times \mathbf{c}|\,\,=\sqrt{15}\](given) \[\Rightarrow \] \[|\mathbf{b}||\mathbf{c}|\sin \alpha =\sqrt{15}\] \[\Rightarrow \] \[\sin \alpha =\frac{\sqrt{15}}{4}\] \[\therefore \] \[\cos \alpha =\frac{1}{4}\] Also, \[\mathbf{b}-2\mathbf{c}=\lambda \mathbf{a}\] \[\Rightarrow \] \[{{(\mathbf{b}-2\mathbf{c})}^{2}}={{\lambda }^{2}}{{(\mathbf{a})}^{2}}\] \[\Rightarrow \] \[|\mathbf{b}{{|}^{2}}+4|\mathbf{c}{{|}^{2}}-4\,\,\mathbf{b}\cdot \mathbf{c}={{\lambda }^{2}}|\mathbf{a}{{|}^{2}}\] \[\Rightarrow \] \[16+4-4\{|\mathbf{b}||\mathbf{c}|\cos \alpha \}={{\lambda }^{2}}\] \[\Rightarrow \] \[{{\lambda }^{2}}=16\] \[\Rightarrow \] \[\lambda =\pm 4\]
You need to login to perform this action.
You will be redirected in
3 sec
