Question

If the vectors $\overline{a},\overline{b},\overline{c}$ are non- coplanar and $l,m,n$ are distinct scalars, then $\left[\right(l\overline{a}+m\overline{b}+n\overline{c}\left)\right(l\overline{b}+m\overline{c}+n\overline{a}\left)\right(l\overline{c}+m\overline{a}+n\overline{b}\left)\right]=0$

• A. $|m+mn+n|=0$
• B. $l+m+n=0$
• C. $l^2+m^2+n^2=0$
• D. $l^3+m^3+n^3=0$

Answer 1

The correct option is B $l+m+n=0$

We have

$\left[\right(l\overline{a}+m\overline{b}+n\overline{c}\left)\right(l\overline{b}+m\overline{c}+n\overline{a}\left)\right(l\overline{c}+m\overline{a}+n\overline{b}\left)\right]=0$

$\Rightarrow \left|\begin{array}{ccc}l& m& n\\ n& l& m\\ m& n& l\end{array}\right|=0$

$\Rightarrow l(l^2-mn)-m(ln-m^2)+n(n^2-lm)=0$

$\Rightarrow l^3+m^3+n^3-3lmn=0$

$\Rightarrow (l+m+n)(l^2+m^2+n^2-lm-mn-ln)=0$

So, $l+m+n=0$