IIT-JEE 1998Question 1
If $\vec a=\hat i+\hat j+\hat k$, $\vec b=4\hat i+3\hat j+4\hat k$ and $\vec c=\hat i+\alpha\hat j+\beta\hat k$ are linearly dependent vectors and $|\vec c|=\sqrt3$, then
Vector Algebra
Solution
$\det[\vec a,\vec b,\vec c]=1-\beta=0$ gives $\beta=1$; $|\vec c|^2=1+\alpha^2+1=3$ gives $\alpha=\pm1$.