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
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
Show 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.
Updated On: Jun 24, 2026
- a parallelogram but not rhombus
- a square
- a quadrilateral which is not a parallelogram
- a rectangle
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Write the position vectors.
\[ A=(7,-4,7),\quad B=(1,-6,10) \] \[ C=(-1,-3,4),\quad D=(5,-1,5) \]
Step 2: Find the side vectors.
\[ \overrightarrow{AB}=B-A \] \[ \overrightarrow{AB}=(1-7,-6+4,10-7) \] \[ \overrightarrow{AB}=(-6,-2,3) \] Also, \[ \overrightarrow{DC}=C-D \] \[ \overrightarrow{DC}=(-1-5,-3+1,4-5) \] \[ \overrightarrow{DC}=(-6,-2,-1) \]
Step 3: Check whether opposite sides are equal.
For \(ABCD\) to be a parallelogram, \[ \overrightarrow{AB}=\overrightarrow{DC} \] But, \[ \overrightarrow{AB}=(-6,-2,3) \] and \[ \overrightarrow{DC}=(-6,-2,-1) \] These are not equal.
Therefore, \(ABCD\) is not a parallelogram.
Step 4: Final conclusion.
Hence, \(ABCD\) is \[ \boxed{\text{a quadrilateral which is not a parallelogram}} \]
\[ A=(7,-4,7),\quad B=(1,-6,10) \] \[ C=(-1,-3,4),\quad D=(5,-1,5) \]
Step 2: Find the side vectors.
\[ \overrightarrow{AB}=B-A \] \[ \overrightarrow{AB}=(1-7,-6+4,10-7) \] \[ \overrightarrow{AB}=(-6,-2,3) \] Also, \[ \overrightarrow{DC}=C-D \] \[ \overrightarrow{DC}=(-1-5,-3+1,4-5) \] \[ \overrightarrow{DC}=(-6,-2,-1) \]
Step 3: Check whether opposite sides are equal.
For \(ABCD\) to be a parallelogram, \[ \overrightarrow{AB}=\overrightarrow{DC} \] But, \[ \overrightarrow{AB}=(-6,-2,3) \] and \[ \overrightarrow{DC}=(-6,-2,-1) \] These are not equal.
Therefore, \(ABCD\) is not a parallelogram.
Step 4: Final conclusion.
Hence, \(ABCD\) is \[ \boxed{\text{a quadrilateral which is not a parallelogram}} \]
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