Question

Given 3 vectors: $\overrightarrow{V_1}=a\hat{i}+b\hat{j}+c\hat{k}$, $\overrightarrow{V_2}=b\hat{i}+c\hat{j}+a\hat{k}$ and $\overrightarrow{V_3}=c\hat{i}+a\hat{j}+b\hat{k}$. In which of the following conditions $\overrightarrow{V_1},\overrightarrow{V_2}$ and $\overrightarrow{V_3}$ are linearly independent?

• A. $a+b+c=0$ and $a^2+b^2+c^2\ne ab+bc+ca$
• B. $a+b+c=0$ and $a^2+b^2+c^2=ab+bc+ca$
• C. $a+b+c\ne 0$ and $a^2+b^2+c^2=ab+bc+ca$
• D. $a+b+c\ne 0$ and $a^2+b^2+c^2\ne ab+bc+ca$

Answer 1

The correct option is D $a+b+c\ne 0$ and $a^2+b^2+c^2\ne ab+bc+ca$

The vectors are linearly independent and so the determinant formed by their co-efficients must be non - zero.

So, $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|\ne 0$

$\therefore abc-a^3-b^3+abc+abc-c^3\ne 0$
$\therefore a^3+b^3+c^3-3abc\ne 0$
$\therefore (a+b+c)\times (a^2+b^2+c^2-ab-bc-ca)\ne 0$