Question

Let $\overline{a}=2\overline{i}-3\overline{j}+\hat{k},\overline{b}=\hat{i}-\hat{j}+3\hat{k}and\overline{c}=\hat{i}+\hat{j}-\hat{k}$ Then the system of vectors $\overline{a},\overline{b}and\overline{c}$

• A. Is linearly dependent
• B. Is linearly independent
• C. Cannot be determined
• D. None of the above

Answer 1

The correct option is B Is linearly independent

To examine the dependence or independence of vectors first form the equation $x\overline{a}+y\overline{b}+z\overline{c}=\overline{0}$, Then find the scalars x, y and z. If there exists a non-zero solution for x, y and z, the system will be linearly dependent. If the only solution comes out to be x = y = z = 0, the system will be linearly independent. Thus, $x\overline{a}+y\overline{b}+z\overline{c}=\overline{0},$
$\Rightarrow x(2\hat{i}-3\hat{j}+\hat{k})+y(\hat{i}-\hat{j}+3\hat{k})+z(\hat{i}+\hat{j}-\hat{k})=0$
Equating the coefficients of $\hat{i},\hat{j}and\hat{k}$
We get,
2x + y + z = 0 - Equating coefficients of$\hat{i}$ (1)
-3x - y + z = 0 - Equating coefficients of $\hat{j}$ (2)
x + 3y - z = 0 - Equating coefficients of $\hat{k}$ (3)
Adding (1) and (3) we get
3x + 4y = 0 ….(4)
Adding (2) and (3) we get
-2x + 2y = 0 ….(5)
or -4x + 4y = 0 ….(6)
Equation (4) – equation (6) gives 7x = 0
$\Rightarrow$ x = 0
Putting x = 0 in (4) we get
y = 0
Put x = y = 0 in (3)
We get
z = 0.
So x = y = z = 0 is the only solution. So the system of the given vectors is linearly independent.