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Vector Algebra question

2003 · Shift 0 · Q97
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Vector Algebra question

2003 · Shift 0 · Q97

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Consider points A,B,CA, B, CA,B,C and DDD with position vectors 7i^−4j^+7k^,i^−6j^+10k^,−i^−3j^+4k^7\widehat i - 4\widehat j + 7\widehat k,\widehat i - 6\widehat j + 10\widehat k, - \widehat i - 3\widehat j + 4\widehat k7i−4j​+7k,i−6j​+10k,−i−3j​+4k and 5i^−j^+5k^5\widehat i - \widehat j + 5\widehat k5i−j​+5k respectively. Then ABCDABCDABCD is a :
  1. A
    parallelogram but not a rhombus
  2. B
    square
  3. C
    rhombus
  4. D
    None
View written solutionFree

Correct answer: D

  1. Write the position vectors as coordinates

Given: A=(7,−4,7),B=(1,−6,10),C=(−1,−3,4),D=(5,−1,5)A=(7,-4,7),\quad B=(1,-6,10),\quad C=(-1,-3,4),\quad D=(5,-1,5)A=(7,−4,7),B=(1,−6,10),C=(−1,−3,4),D=(5,−1,5)

We must determine the nature of quadrilateral ABCDABCDABCD.

  1. Find the side vectors
-6-(-4), 10-7)=(-6,-2,3)$$ $$\overrightarrow{BC}=C-B=(-1-1, -3-(-6), 4-10)=(-2,3,-6)$$ $$\overrightarrow{CD}=D-C=(5-(-1), -1-(-3), 5-4)=(6,2,1)$$ $$\overrightarrow{DA}=A-D=(7-5, -4-(-1), 7-5)=(2,-3,2)$$ --- 3. **Check whether opposite sides are parallel** For a parallelogram, we need $$\overrightarrow{AB}\parallel \overrightarrow{DC} \quad \text{and} \quad \overrightarrow{BC}\parallel \overrightarrow{AD}$$ Now, $$\overrightarrow{DC}=C-D=(-1-5,-3-(-1),4-5)=(-6,-2,-1)$$ Compare with $$\overrightarrow{AB}=(-6,-2,3)$$ These are **not** scalar multiples, so $AB \not\parallel DC$. Also, $$\overrightarrow{AD}=D-A=(5-7,-1-(-4),5-7)=(-2,3,-2)$$ Compare with $$\overrightarrow{BC}=(-2,3,-6)$$ These are also **not** scalar multiples, so $BC \not\parallel AD$. Hence, $ABCD$ is **not a parallelogram**. So options A, B, C are already ruled out. --- 4. **Optional check using diagonals** In a parallelogram, diagonals bisect each other. Midpoint of $AC$: $$M_{AC}=\left(\frac{7+(-1)}2,\frac{-4+(-3)}2,\frac{7+4}2\right)=\left(3,-\frac72,\frac{11}2\right)$$ Midpoint of $BD$: $$M_{BD}=\left(\frac{1+5}2,\frac{-6+(-1)}2,\frac{10+5}2\right)=\left(3,-\frac72,\frac{15}2\right)$$ These are not equal, confirming again that $ABCD$ is not a parallelogram. --- 5. **Conclusion** Since $ABCD$ is not even a parallelogram, it cannot be a rhombus or square. Therefore, the correct option is: $$\boxed{\text{D: None}}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: **D** Our derived answer: **D** They agree.
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