Vector Algebra Question 184
Question: If b and c are any two non-collinear unit vectors and a is any vector, then $ (\mathbf{a},.,\mathbf{b}),\mathbf{b}+(\mathbf{a},.,\mathbf{c}),\mathbf{c}+\frac{\mathbf{a},.,(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|},(\mathbf{b}\times \mathbf{c})= $
[IIT 1996]
Options:
A) a
B) b
C) c
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
- Let $ \mathbf{i} $ be a unit vector in the direction of $ \mathbf{b},,\mathbf{j} $ in the direction of $ \mathbf{c}. $ Note that $ \mathbf{b}=\mathbf{i} $ and $ \mathbf{c}=\mathbf{j} $
We have $ \mathbf{b}\times \mathbf{c}=,|\mathbf{b}||\mathbf{c}|\sin \alpha ,\mathbf{k}=\sin \alpha ,\mathbf{k} $ , where $ \mathbf{k} $ is a unit vector perpendicular to $ \mathbf{b} $ and $ \mathbf{c}. $
$ \Rightarrow ,|\mathbf{b}\times \mathbf{c}|,=\sin \alpha \Rightarrow \mathbf{k}=\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|} $ Any vector $ \mathbf{a} $ can be written as a linear combination of $ \mathbf{i},\mathbf{j} $ and $ \mathbf{k}. $
Let $ \mathbf{a}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}. $
Now $ \mathbf{a},.,\mathbf{b}=\mathbf{a},.,\mathbf{i}=a_1, $ $ \mathbf{a},.,\mathbf{c}=\mathbf{a},.,\mathbf{j}=a_2 $ and $ \mathbf{a},.\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|}=\mathbf{a},.,\mathbf{k}=a_3 $
Thus $ (\mathbf{a},.,\mathbf{b})\mathbf{b}+(\mathbf{a},.,\mathbf{c})\mathbf{c}+\frac{\mathbf{a},.,(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}(\mathbf{b}\times \mathbf{c}| $
$ =a_1\mathbf{b}+a_2\mathbf{c}+a_3\frac{(\mathbf{b}\times \mathbf{c})}{|\mathbf{b}\times \mathbf{c}|}=a_1\mathbf{i}+a_2\mathbf{j}+a_3\mathbf{k}=\mathbf{a} $ .
BETA
