JEE Main 2020 (Online) 9th January Morning ShiftQuestion 1
If the vectors, $\vec p=(a+1)\hat i+a\hat j+a\hat k$, $\vec q=a\hat i+(a+1)\hat j+a\hat k$ and $\vec r=a\hat i+a\hat j+(a+1)\hat k$ ($a\in\mathbf{R}$) are coplanar and $3(\vec p\cdot\vec q)^2-\lambda|\vec r\times\vec q|^2=0$, then the value of $\lambda$ is ________.
Vector Algebra
Answer$1$
Solution
Coplanar: $\det(aJ+I)=1+3a=0\Rightarrow a=-\frac13$. $\vec p=\frac13(2,-1,-1),\ \vec q=\frac13(-1,2,-1),\ \vec r=\frac13(-1,-1,2)$. $\vec p\cdot\vec q=-\frac13$, $\vec r\cdot\vec q=-\frac13$, $|\vec r|^2=|\vec q|^2=\frac23$. $|\vec r\times\vec q|^2=\frac49-\frac19=\frac13$. $3(\vec p\cdot\vec q)^2=\frac13=\lambda\cdot\frac13\Rightarrow\lambda=1$.