11.0 Linearly Dependent and Independent Vectors

A set of vectors is said to be linearly dependent if one of the vectors in the set can be written as a linear combination of the others.

A set of vectors ${a_1},{a_2},{a_{3...........}},{a_n}$ is said to be linearly dependent, if there exists some scalars ${k_1},{k_2},{k_{3...........}},{k_n}$ (not all zero) such that $$\overrightarrow {{a_1}} {k_1} + \overrightarrow {{a_2}} {k_2} + \overrightarrow {{a_3}} {k_{3...........}}+\overrightarrow {{a_n}} {k_n} = 0$$

A set of vector is said to be linearly independent if no vector in the set can be written as a linear combination of the others.

A set of vectors ${a_1},{a_2},{a_{3...........}},{a_n}$ is said to be linearly independent, if every relation of the type

$$\overrightarrow {{a_1}} {k_1} + \overrightarrow {{a_2}} {k_2} + \overrightarrow {{a_3}} {k_{3...........}}+\overrightarrow {{a_n}} {k_n} = 0$$for all ${k_1} = {k_2} = {k_3} = ....... = {k_n} = 0$

1) Two non-zero, non-collinear vectors are linearly independent.

2) Any two collinear vectors are linearly dependent.

3) Any three non coplanar vectors are linearly independent.

4) Any three coplanar vectors are linearly dependent.

5) Any four vectors in $3-$dimensional space are linearly dependent.

Question 19. If $\overrightarrow a = \widehat i + \widehat j + \widehat k$, $\overrightarrow b = 4\widehat i + 3\widehat j + 4\widehat k$ and $\overrightarrow c = \widehat i + \alpha \widehat j + \beta \widehat k$ are linearly dependent vectors and $\left| {\overrightarrow c } \right| = \sqrt 3 $, then find the value of $\alpha$ and $\beta$.

Solution: If $\overrightarrow a $, $\overrightarrow b $ and $\overrightarrow c $ are linearly dependent vectors, we can say that $\overrightarrow c $ can be written as a linear combination of $\overrightarrow a $ and $\overrightarrow b $ i.e., $$\begin{equation} \begin{aligned} \overrightarrow c = x\overrightarrow a + y\overrightarrow b \\ \widehat i + \alpha \widehat j + \beta \widehat k = x(\hat i + \hat j + \hat k) + y(4\hat i + 3\hat j + 4\hat k) \\\end{aligned} \end{equation} $$ Equating coefficients of ${\hat i}$, ${\hat j}$ and ${\hat k}$, we get $$1 = x + 4y;\quad \alpha = x + 3y;\quad \beta = x + 4y$$ On solving the equations, we get $$\beta = 1$$ Now, it is given that $\left| {\overrightarrow c } \right| = \sqrt 3 $, $$\begin{equation} \begin{aligned} \left| c \right| = 1 + {\alpha ^2} + {\beta ^2} = 3 \\ \Rightarrow 1 + {\alpha ^2} + 1 = 3 \\ \Rightarrow \alpha = \pm 1 \\\end{aligned} \end{equation} $$