Vector Algebra Question 363
Question: If the vector $ \mathbf{i}+\mathbf{j}+\mathbf{k} $ makes angles $ \alpha ,,\beta ,,\gamma $ with vectors $ \mathbf{i},,\mathbf{j},\mathbf{k} $ respectively, then
Options:
A) $ \alpha =\beta \ne \gamma $
B) $ \alpha =\gamma \ne \beta $
C) $ \beta =\gamma \ne \alpha $
D) $ \alpha =\beta =\gamma $
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Answer:
Correct Answer: D
Solution:
- Angle between $ \mathbf{i}+\mathbf{j}+\mathbf{k} $ and $ \mathbf{i} $ is equal to $ {{\cos }^{-1}}{ \frac{(\mathbf{i}+\mathbf{j}+\mathbf{k}),.,\mathbf{i}}{|\mathbf{i}+\mathbf{j}+\mathbf{k}||\mathbf{i}|} }\Rightarrow \alpha ={{\cos }^{-1}}( \frac{1}{\sqrt{3}} ) $ Similarly angle between $ \mathbf{i}+\mathbf{j}+\mathbf{k} $ and $ \mathbf{j} $ is $ \beta ={{\cos }^{-1}}( \frac{1}{\sqrt{3}} ) $ and between $ \mathbf{i}+\mathbf{j}+\mathbf{k} $ and $ \mathbf{k} $ is $ \gamma ={{\cos }^{-1}}( \frac{1}{\sqrt{3}} ),. $ Hence $ \alpha =\beta =\gamma . $
BETA
