JEE Main & Advanced
Mathematics
Vector Algebra
Question Bank
Critical Thinking
question_answer
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]
A)a
B)b
C)c
D)0
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 \[\mathbf{a}\,.\frac{\mathbf{b}\times \mathbf{c}}{|\mathbf{b}\times \mathbf{c}|}=\mathbf{a}\,.\,\mathbf{k}={{a}_{3}}\]