Vector Algebra Question 237
Question: A vector r is equally inclined with the co-ordinate axes. If the tip of r is in the positive octant and |r| = 6, then $ \mathbf{r} $ is
Options:
A) $ 2\sqrt{3}(\mathbf{i}-\mathbf{j}+\mathbf{k}) $
B) $ 2\sqrt{3}(-\mathbf{i}+\mathbf{j}+\mathbf{k}) $
C) $ 2\sqrt{3}(\mathbf{i}+\mathbf{j}-\mathbf{k}) $
D) $ 2\sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k}) $
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Answer:
Correct Answer: D
Solution:
- Let $ l,m,n $ be the d.c’s of $ \mathbf{r}. $ Then $ l=m=n $ , (given) \ $ l^{2}+m^{2}+n^{2}=1\Rightarrow 3l^{2}=1\Rightarrow l=\frac{1}{\sqrt{3}}=m=n $ Now, $ \mathbf{r}=|\mathbf{r}|(l\mathbf{i}+m\mathbf{j}+n\mathbf{k})=6( \frac{1}{\sqrt{3}}\mathbf{i}+\frac{1}{\sqrt{3}}\mathbf{j}+\frac{1}{\sqrt{3}}\mathbf{k} ) $ Hence, $ \mathbf{r}=2\sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k}) $ .
BETA
