Vector Algebra Question 65
Question: Let $ \alpha ,\beta ,\gamma $ be distinct real numbers. The points with position vectors $ \alpha \mathbf{i}+\beta \mathbf{j}+\gamma \mathbf{k},\beta \mathbf{i}+\gamma \mathbf{j}+\alpha \mathbf{k},\gamma \mathbf{i}+\alpha \mathbf{j}+\beta \mathbf{k} $
[IIT Screening 1994]
Options:
A) Are collinear
B) Form an equilateral triangle
C) Form a scalene triangle
D) Form a right angled triangle
Show Answer
Answer:
Correct Answer: B
Solution:
- Let $ P,,Q $ and $ R $ be points having position vectors $ \alpha ,\mathbf{i}+\beta ,\mathbf{j}+\gamma ,\mathbf{k}, $ $ \beta ,\mathbf{i}+\gamma ,\mathbf{j}+\alpha ,\mathbf{k} $ and $ \gamma ,\mathbf{i}+\alpha j+\beta \mathbf{k} $ respectively. Then, $ |\overrightarrow{PQ}|,=,|\overrightarrow{QR}|,=,|\overrightarrow{RP}|,=\sqrt{{{(\alpha -\beta )}^{2}}+{{(\beta -\gamma )}^{2}}+{{(\gamma -\alpha )}^{2}}} $ Hence $ \Delta PQR $ is an equilateral triangle.
BETA
