Question

If $a,b,c$ are distinct positive real numbers then the vectors $(a,b,c)$ , $(b,c,a)$ and $(c,a,b)$ are

• A. Coplanar
• B. Non coplanar
• C. Collinear
• D. $0$

Answer 1

Answer: (B)

Non coplanar

Let
$\overrightarrow{A}=a\hat{i}+b\hat{j}+c\hat{k}$; $\overrightarrow{B}=b\hat{i}+c\hat{j}+a\hat{k}$ and $\overrightarrow{C}=c\hat{i}+a\hat{j}+b\hat{k}$.

Then
$\overrightarrow{A}\times \overrightarrow{B}=\left(\right(ac)k-a^2\hat{j})+(-(b^2)k+ab\hat{i})+\left(\right(bc)j-c^2\hat{i})$

$=\hat{i}(ab-c^2)+\hat{j}(bc-a^2)+\hat{k}(ac-b^2)$

Now
$(\overrightarrow{A}\times \overrightarrow{B})\cdot \overrightarrow{C}$

$=\left[\hat{i}\right(ab-c^2)+\hat{j}(bc-a^2)+\hat{k}(ac-b^2\left)\right].(c\hat{i}+a\hat{j}+b\hat{k})$

$=(ab-c^2)c+(bc-a^2)a+(ac-b^2)b$

$=abc-c^3+abc-a^3+abc-b^3$

$=3abc-(a^3+b^3+c^3)$

Since all $a,b,c>0$ and $a\ne b\ne c$ hence

$3abc-(a^3+b^3+c^3)\ne 0$.

Therefore
$(\overrightarrow{A}\times \overrightarrow{B})\cdot \overrightarrow{C}\ne 0$.

Therefore, the vectors are non-coplanar.