Vector Algebra Question 250
Question: The direction cosines of the resultant of the vectors $ (\mathbf{i}+\mathbf{j}+\mathbf{k}), $ $ (-\mathbf{i}+\mathbf{j}+\mathbf{k}), $ $ (\mathbf{i}-\mathbf{j}+\mathbf{k}) $ and $ (\mathbf{i}+\mathbf{j}-\mathbf{k}), $ are
Options:
A) $ ( \frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{6}} ) $
B) $ ( \frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}} ) $
C) $ ( -\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}},,-,\frac{1}{\sqrt{6}} ) $
D) $ ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},,\frac{1}{\sqrt{3}} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
- Resultant vector $ =2\mathbf{i}+2\mathbf{j}+2\mathbf{k}. $ Direction cosines are $ ( \frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}} ),. $
BETA
