Vector Algebra Question 5
Question: The figure formed by the four points $\hat{i}+\hat{j}-\hat{k}, 2 \hat{i}+3 \hat{j}, 3 \hat{i}+5 \hat{j}-2 \hat{k}$ and $\hat{k}-\hat{j}$ is
[MP PET 2004]
Options:
A) Rectangle
B) Parallelogram
C) Trapezium
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- Let $ A\equiv (1, 1,-1),B\equiv (2, 3, 0),C\equiv (3, 5, -2), D\equiv (0, -1, 1) $ So, $ \overrightarrow{AB}=(1, 2, 1),\overrightarrow{BC}=(1, 2,-2),\overrightarrow{CD}=(-3,-6, 3),\overrightarrow{DA}=(1, 2, -2) $ Clearly, $ \overrightarrow{BC}||\overrightarrow{DA}, $ but $ AB\ne CD $ So, it is a trapezium.
BETA
