A) \[-10\hat{i}+5\hat{j}\]
B) \[-14\hat{i}+5\hat{j}\]
C) \[-14\hat{i}-5\hat{j}\]
D) \[-10\hat{i}-5\hat{j}\]
Correct Answer: A
Solution :
Since \[\vec{a},\vec{b}\]and \[\vec{c}\]are coplanar vectors. \[\therefore \]\[\left[ \begin{matrix} {\vec{a}} & {\vec{b}} & {\vec{c}} \\ \end{matrix} \right]=0\] \[\Rightarrow \]\[\left| \begin{matrix} 1 & 2 & 4 \\ 1 & \lambda & 4 \\ 2 & 4 & {{\lambda }^{2}}-1 \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1[\lambda ({{\lambda }^{2}}-1)-16]-2({{\lambda }^{2}}-1-8)+4(4-2\lambda )=0\] \[\Rightarrow \]\[{{\lambda }^{3}}-\lambda -16-2{{\lambda }^{2}}+18+16-8\lambda =0\] \[\Rightarrow \]\[{{\lambda }^{3}}-2{{\lambda }^{2}}-9\lambda +18=0\] \[\Rightarrow \]\[{{\lambda }^{2}}(\lambda -2)-9(\lambda -2)=0\] \[\Rightarrow \]\[(\lambda +3)(\lambda -3)(\lambda -2)=0\] \[\Rightarrow \]\[\lambda =3,-3,2\] When \[\lambda =\pm 3,\]then a is parallel to \[\vec{c}\]. When\[\lambda =2,\vec{a}=\hat{i}+2\hat{j}+4\hat{k}\]and\[\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}\] \[\therefore \]\[\vec{a}\times \vec{c}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & 4 \\ 2 & 4 & 3 \\ \end{matrix} \right|=-10\hat{i}+5\hat{j}\]
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