Question

If the position vectors of the points \(A,B,C,D\) are \[ 7\hat{i}-4\hat{j}+7\hat{k},\quad \hat{i}-6\hat{j}+10\hat{k},\quad -\hat{i}-3\hat{j}+4\hat{k},\quad 5\hat{i}-\hat{j}+5\hat{k} \] respectively, then \(ABCD\) is

• A. a parallelogram but not rhombus
• B. a square
• C. a quadrilateral which is not a parallelogram
• D. a rectangle

Hint:

To check whether four points form a parallelogram, compare opposite side vectors. If \[ \overrightarrow{AB}=\overrightarrow{DC} \] and \[ \overrightarrow{BC}=\overrightarrow{AD}, \] then the quadrilateral is a parallelogram.

Answer 1

The Correct Option is C

Step 1: Compute $\overrightarrow{AB}$ and $\overrightarrow{DC}$.
$A=(7,-4,7)$, $B=(1,-6,10)$, $C=(-1,-3,4)$, $D=(5,-1,5)$. \[ \overrightarrow{AB} = B-A = (-6,-2,3) \] \[ \overrightarrow{DC} = C-D = (-6,-2,-1) \]

Step 2: Check if opposite sides are equal.
For a parallelogram $ABCD$, we need $\overrightarrow{AB} = \overrightarrow{DC}$. But $(-6,-2,3) \neq (-6,-2,-1)$ (third components differ). So $ABCD$ is NOT a parallelogram.

Step 3: Check $\overrightarrow{AD}$ and $\overrightarrow{BC}$ as well.
$\overrightarrow{AD} = D-A = (-2,3,-2)$ and $\overrightarrow{BC} = C-B = (-2,3,-6)$. These are also unequal. So no pair of opposite sides is parallel and equal.

Step 4: Conclude the shape.
Since neither pair of opposite sides satisfies the parallelogram condition, $ABCD$ is a general quadrilateral that is not a parallelogram.

Step 5: Check if it could be special in another way.
Without parallel sides, it cannot be a rhombus, rectangle, or square either.

Step 6: State the answer.
\[ \boxed{\text{a quadrilateral which is not a parallelogram}} \]