Question

If \[\vec a\], $\vec b$ and $\vec c$ are linearly independent vectors, then which one of the following sets of vectors is linearly dependent?
A. $\vec a + \vec b$, $\vec b + \vec c$, $\vec c + \vec a$
B. $\vec a - \vec b$, $\vec b - \vec c$, $\vec c - \vec a$
C. $\vec a \times \vec b$, \[\vec b \times \vec c\], $\vec c \times \vec a$
D. $\vec a + 2\vec b + 3\vec c$, $\vec b - \vec c + \vec a$, $\vec a + \vec c$

Answer 1

Hint:We must know what we mean by linearly independent vectors. Linearly independent vectors are those which do not lie on the same plane, that is, the vectors are not coplanar. Three vectors\[\vec a\], $\vec b$ and $\vec c$ are said to be linearly independent if $x\vec a + y\vec b + z\vec c = 0$ holds true only for $x = y = z = 0$. If there is any other solution to the equation $x\vec a + y\vec b + z\vec c = 0$, then the c=vectors are said to be linearly dependent. We can solve the given problem by analysing the given options.

Complete step by step answer:
In option (A), we have the vectors $\vec a + \vec b$, $\vec b + \vec c$, $\vec c + \vec a$.
$\vec a + \vec b$, $\vec b + \vec c$, $\vec c + \vec a$ can be written in the form,
$ \Rightarrow x\left( {\vec a + \vec b} \right) + y\left( {\vec b + \vec c} \right) + z\left( {\vec c + \vec a} \right) = 0$
$ \Rightarrow \vec a\left( {x + z} \right) + \vec b\left( {x + y} \right) + \vec c\left( {y + z} \right) = 0 - - - - - \left( 1 \right)$
But, it is given that \[\vec a\], $\vec b$ and $\vec c$ are linearly independent.
So, the coefficients of \[\vec a\], $\vec b$ and $\vec c$ must be equal to zero.
Therefore, for equation $\left( 1 \right)$ to be true, we have,
$x + z = 0$
$\Rightarrow x + y = 0$
$\Rightarrow y + z = 0$
Now, we can clearly see that the above three conditions are satisfied if and only if,
$x = y = z = 0$
Hence, the coefficients of the vectors $\vec a + \vec b$, $\vec b + \vec c$, $\vec c + \vec a$ in the equation $x\left( {\vec a + \vec b} \right) + y\left( {\vec b + \vec c} \right) + z\left( {\vec c + \vec a} \right) = 0$ are all equal to zero.
Hence, we can conclude that vectors $\vec a + \vec b$, $\vec b + \vec c$, $\vec c + \vec a$ are linearly independent.

In option (B), we have the vectors $\vec a - \vec b$, $\vec b - \vec c$, $\vec c - \vec a$.
$\vec a - \vec b$, $\vec b - \vec c$, $\vec c - \vec a$ can be written in the form,
$ \Rightarrow x\left( {\vec a - \vec b} \right) + y\left( {\vec b - \vec c} \right) + z\left( {\vec c - \vec a} \right) = 0$
$ \Rightarrow \vec a\left( {x - z} \right) + \vec b\left( {y - x} \right) + \vec c\left( {z - y} \right) = 0 - - - - - \left( 2 \right)$
But, it is given that \[\vec a\], $\vec b$ and $\vec c$ are linearly independent.
So, the coefficients of \[\vec a\], $\vec b$ and $\vec c$ must be equal to zero.
Therefore, for equation $\left( 2 \right)$ to be true, we have,
$x - z = 0$
$\Rightarrow y - x = 0$
$\Rightarrow z - y = 0$
Now, we can clearly see that the above three conditions are satisfied if and only if,
$x = y = z$
So, x, y and z can assume any real value.
Hence, the coefficients of the vectors $\vec a - \vec b$, $\vec b - \vec c$, $\vec c - \vec a$ in the equation $x\left( {\vec a - \vec b} \right) + y\left( {\vec b - \vec c} \right) + z\left( {\vec c - \vec a} \right) = 0$ are not always equal to zero.
Hence, we can conclude that vectors $\vec a - \vec b$, $\vec b - \vec c$, $\vec c - \vec a$ are linearly dependent.

In option (C), we have the vectors $\vec a \times \vec b$, \[\vec b \times \vec c\], $\vec c \times \vec a$.
$\vec a \times \vec b$, \[\vec b \times \vec c\], $\vec c \times \vec a$ are three vectors which are pairwise cross products of \[\vec a\], $\vec b$ and $\vec c$.
Now, the cross product of two vectors always directs towards the perpendicular direction of the two vectors.
Therefore, $\vec a \times \vec b$ vector is perpendicular to both \[\vec a\] and $\vec b$.
Similarly, \[\vec b \times \vec c\] vector is perpendicular to $\vec b$ and $\vec c$.
And, $\vec c \times \vec a$ vector is perpendicular to $\vec c$ and \[\vec a\].
Now, since, \[\vec a\], $\vec b$ and $\vec c$are linearly independent, so the vectors perpendicular to them will also be linearly independent as these three vectors will also lie in the same plane. Therefore, $\vec a \times \vec b$, \[\vec b \times \vec c\], $\vec c \times \vec a$ are also linearly independent.

In option (D), we have the vectors $\vec a + 2\vec b + 3\vec c$, $\vec b - \vec c + \vec a$, $\vec a + \vec c$.
$\vec a + 2\vec b + 3\vec c$, $\vec b - \vec c + \vec a$, $\vec a + \vec c$ can be written in the form,
$ \Rightarrow x\left( {\vec a + 2\vec b + 3\vec c} \right) + y\left( {\vec b - \vec c + \vec a} \right) + z\left( {\vec a + \vec c} \right) = 0$
$ \Rightarrow \vec a\left( {x + y + z} \right) + \vec b\left( {2x + y} \right) + \vec c\left( {3x - y + z} \right) = 0 - - - - - \left( 3 \right)$
But, it is given that \[\vec a\], $\vec b$ and $\vec c$ are linearly independent.
So, the coefficients of \[\vec a\], $\vec b$ and $\vec c$ must be equal to zero.
Therefore, for equation $\left( 3 \right)$ to be true, we have,
$x + y + z = 0$
$\Rightarrow 2x + y = 0$
$\Rightarrow 3x - y + z = 0$
Now, we have to solve the three equations above to find the values of x, y and z.
So, substituting the value of y from $2x + y = 0$ into $x + y + z = 0$, we get,
$ \Rightarrow x + \left( { - 2x} \right) + z = 0$
\[ \Rightarrow z - x = 0 - - - - \left( 4 \right)\]
Similarly, substituting the value of y in $3x - y + z = 0$, we get,
$ \Rightarrow 3x - 2x + z = 0$
$ \Rightarrow x - z = 0 - - - - \left( 5 \right)$
Now, adding equations $\left( 4 \right)$ and $\left( 5 \right)$, we get,
$ \Rightarrow 2x = 0$
$ \Rightarrow x = 0$
So, putting the value of x in equation $\left( 4 \right)$, we get the value of z as $0$.
Putting value of x in equation $\left( 2 \right)$, we get,
$ \Rightarrow 2\left( 0 \right) + y = 0$
$ \Rightarrow y = 0$
Hence, the coefficients of the vectors $\vec a + 2\vec b + 3\vec c$, $\vec b - \vec c + \vec a$, $\vec a + \vec c$ in the equation $x\left( {\vec a + 2\vec b + 3\vec c} \right) + y\left( {\vec b - \vec c + \vec a} \right) + z\left( {\vec a + \vec c} \right) = 0$ are all equal to zero.

Hence, we can conclude that vectors $\vec a + 2\vec b + 3\vec c$, $\vec b - \vec c + \vec a$, $\vec a + \vec c$ are linearly independent.

Note:To easily find such answers to such questions with options, it is always preferable and easier to use the option analysis method. This is because it leads to less confusion and errors and saves a lot of time in competitive examinations. We must understand the meaning of the term ‘linearly dependent vectors’ and ‘linearly independent vectors’ in order to tackle the given problem.