JEE Main & Advanced Mathematics Vector Algebra Linear Combination of Vectors

A vector \[\mathbf{r}\] is said to be a linear combination of vectors \[\mathbf{a},\,\mathbf{b},\,\mathbf{c}.....\] etc, if there exist scalars \[x,y,z\] etc., such that \[\mathbf{r}=x\mathbf{a}+y\mathbf{b}+z\mathbf{c}+....\]

Examples : Vectors \[{{\mathbf{r}}_{\text{1}}}=2\mathbf{a}+\mathbf{b}+3\mathbf{c},\,{{\mathbf{r}}_{2}}=\mathbf{a}+3\mathbf{b}+\sqrt{2}\mathbf{c}\] are linear combinations of the vectors \[\mathbf{a},\,\mathbf{b},\,\mathbf{c}\].            

(1) Collinear and Non-collinear vectors : Let \[\mathbf{a}\] and \[\mathbf{b}\] be two collinear vectors and let \[\mathbf{x}\] be the unit vector in the direction of \[\mathbf{a}\]. Then the unit vector in the direction of \[\mathbf{b}\] is \[\mathbf{x}\] or \[-\mathbf{x}\] according as \[\mathbf{a}\] and \[\mathbf{b}\] are like or unlike parallel vectors. Now, \[\mathbf{a}=|\mathbf{a}|\mathbf{\hat{x}}\]\[\mathbf{a}\times (\mathbf{b}\times \mathbf{c})\ne (\mathbf{a}\times \mathbf{b})\times \mathbf{c}\] and \[\mathbf{b}=\pm |\mathbf{b}|\mathbf{\hat{x}}\].

\[\therefore \]\[\mathbf{a}=\,\left( \frac{|\mathbf{a}|}{|\mathbf{b}|} \right)\,|\mathbf{b}|\mathbf{\hat{x}}\] \[\Rightarrow \] \[\mathbf{a}=\left( \pm \frac{|\mathbf{a}|}{|\mathbf{b}|} \right)\,\mathbf{b}\]\[\Rightarrow \] \[\mathbf{a}=\lambda \mathbf{b}\],  where \[\lambda =\pm \frac{|\mathbf{a}|}{|\mathbf{b}|}\]. Thus, if \[\mathbf{a},\,\mathbf{b}\] are collinear vectors, then \[\mathbf{a}=\lambda \,\mathbf{b}\] or \[\mathbf{b}=\lambda \,\mathbf{a}\] for some scalar \[\lambda \].

(2) Relation between two parallel vectors

(i) If \[\mathbf{a}\] and \[\mathbf{b}\] be two parallel vectors, then there exists a scalar k such that \[\mathbf{a}=k\,\mathbf{b}\] i.e., there exist two non-zero scalar quantities \[x\] and \[y\] so that \[x\,\mathbf{a}+y\,\mathbf{b}=\mathbf{0}\].

If \[\mathbf{a}\] and  \[\mathbf{b}\] be two non-zero, non-parallel vectors then \[x\mathbf{a}+y\mathbf{b}=\mathbf{0}\] \[\Rightarrow \] \[x=0\] and \[y=0\].

Obviously  \[x\mathbf{a}+y\mathbf{b}=\mathbf{0}\] \[\Rightarrow \]\[\left\{ \begin{matrix} \mathbf{a}=\mathbf{0},\,\mathbf{b}=\mathbf{0}  \\ \text{or}  \\ x=\text{0,}\,y=\text{0}  \\ \text{or}  \\ \mathbf{a}||\mathbf{b}  \\ \end{matrix} \right.\]

(ii) If \[\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}\] and \[\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}\] then from the property of parallel vectors, we have \[\mathbf{a}||\mathbf{b}\Rightarrow \frac{{{a}_{1}}}{{{b}_{1}}}=\frac{{{a}_{2}}}{{{b}_{2}}}=\frac{{{a}_{3}}}{{{b}_{3}}}\].

(3) Test of collinearity of three points : Three points with position vectors \[\mathbf{a},\,\mathbf{b,}\,\mathbf{c}\] are collinear iff there exist scalars \[x,y,z\] not all zero such that \[x\mathbf{a}+y\mathbf{b}+z\mathbf{c}=\mathbf{0}\], where \[x+y+z=0\]. If \[\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}\], \[\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}\] and \[\mathbf{c}={{c}_{1}}\mathbf{i}+{{c}_{2}}\mathbf{j}\], then the points with position vector \[\mathbf{a},\,\mathbf{b},\,\mathbf{c}\] will be collinear iff \[\left| \,\begin{matrix} {{a}_{1}} & {{a}_{2}} & 1  \\ {{b}_{1}} & {{b}_{2}} & 1  \\ {{c}_{1}} & {{c}_{2}} & 1  \\ \end{matrix}\, \right|\,=0\].

(4) Test of coplanarity of three vectors : Let \[\mathbf{a}\] and \[\mathbf{b}\] two given non-zero non-collinear vectors. Then any vectors \[\mathbf{r}\] coplanar with \[\mathbf{a}\] and \[\mathbf{b}\] can be uniquely expressed as \[\mathbf{r}=x\mathbf{a}+y\mathbf{b}\] for some scalars \[x\] and \[y\].

(5) Test of coplanarity of Four points : Four points with position vectors \[\mathbf{a},\,\mathbf{b},\,\mathbf{c},\,\mathbf{d}\] are coplanar iff there exist scalars \[x,y,z,u\] not all zero such that \[x\mathbf{a}+y\mathbf{b}+z\mathbf{c}+u\mathbf{d}=\mathbf{0}\], where \[x+y+z+u=0\].

Four points with position vectors

\[\mathbf{a}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k}\],\[\mathbf{b}={{b}_{1}}\mathbf{i}+{{b}_{2}}\mathbf{j}+{{b}_{3}}\mathbf{k}\], \[\mathbf{c}={{c}_{1}}\mathbf{i}+{{c}_{2}}\mathbf{j}+{{c}_{3}}\mathbf{k}\], \[\mathbf{d}={{d}_{1}}\mathbf{i}+{{d}_{2}}\mathbf{j}+{{d}_{3}}\mathbf{k}\] will be coplanar, iff \[\left| \,\begin{matrix} {{a}_{1}} & {{a}_{2}} & {{a}_{3}} & 1  \\ {{b}_{1}} & {{b}_{2}} & {{b}_{3}} & 1  \\ {{c}_{1}} & {{c}_{2}} & {{c}_{3}} & 1  \\ {{d}_{1}} & {{d}_{2}} & {{d}_{3}} & 1  \\ \end{matrix}\, \right|\,=0\].