AIEEE 2003 (Offline)Question 1
Consider points $A,B,C$ and $D$ with position vectors $7\hat i-4\hat j+7\hat k$, $\hat i-6\hat j+10\hat k$, $-\hat i-3\hat j+4\hat k$ and $5\hat i-\hat j+5\hat k$ respectively. Then $ABCD$ is a :
Vector Algebra
Solution
$\overrightarrow{AB}=(-6,-2,3)$, $\overrightarrow{DC}=(-6,-2,-1)$; they are not equal, so $ABCD$ is not a parallelogram (so neither rhombus nor square).