Mathematics - Vector Question with Solution | TestHub

$\because \bar{a},\bar{b}$ and $\bar{c}$ are coplanar $\therefore \left|\begin{array}{ccc}1& 2& 4\\ 1& \lambda & 4\\ 2& 4& \left({\lambda }^2-1\right)\end{array}\right|=0$ $\Rightarrow {\lambda }^3-\lambda -16+2\left(8-{\lambda }^2+1\right)+4(4-2\lambda )=0$ $\Rightarrow {\lambda }^3-2{\lambda }^2-9\lambda +18=0$ i.e. $(\lambda -2)(\lambda -3)(\lambda +3)=0$ For $\lambda =2,\vec{c}=2\hat{i}+4\hat{j}+3\hat{k}$ $\therefore \vec{a}\times \vec{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 4\\ 2& 4& 3\end{array}\right|=-10\hat{i}+5\hat{j}$ For $\lambda =3$ or $-3,\bar{c}=2\bar{a}\Rightarrow \bar{a}\times \bar{c}=0$ (Rejected)