JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
Let the vectors $\vec{u}_1=\hat{i}+\hat{j}+a\hat{k}$, $\vec{u}_2=\hat{i}+b\hat{j}+\hat{k}$ and $\vec{u}_3=c\hat{i}+\hat{j}+\hat{k}$ be coplanar. If the vectors $\vec{v}_1=(a+b)\hat{i}+c\hat{j}+c\hat{k}$, $\vec{v}_2=a\hat{i}+(b+c)\hat{j}+a\hat{k}$ and $\vec{v}_3=b\hat{i}+b\hat{j}+(c+a)\hat{k}$ are also coplanar, then $6(a+b+c)$ is equal to
Vector Algebra
Solution
$\vec u$'s coplanar: $abc-a-b-c+2=0$. $\vec v$'s coplanar: $\det=4abc=0\Rightarrow abc=0$. Then $a+b+c=2$, so $6(a+b+c)=12$. Option (D).