Question

If the vectors $(-bc,b^2+bc,c^2+bc),(a^2+ac,-ac,c^2+ac)$ and $(a^2+ab,b^2+ab,-ab)$ are coplanar (where none of a, b or c is zero). Then?

• A. $a^2+b^2+c^2=1$
• B. $a+b+c=0$
• C. $ab+bc+ca=0$
• D. $a^2+b^2+c^2=(a+b+c{)}^2$

Answer 1

The correct option is C $ab+bc+ca=0$

$\text{solve:- we have to Given three}$

$\text{vectors which are coplaner}$

$\text{so, their scaler triple product}$

$\text{must be zero.}$

$\Rightarrow$

$\left|\begin{array}{ccc}-bc& b^2+bc& c^2+bc\\ a^2+ac& -ac& c^2+ac\\ a^2+ab& b+ab& -ab\end{array}\right|=0$

$R_1\to a{R}_1{R}_2\to b{R}_2,R_3\to C{R}_3$

$\Rightarrow \frac{1}{abc}\left|\begin{array}{ccc}-abc& a(b^2+bc)& a(c^2+bc)\\ b(a^2+ac)& -abc& b(c^2+ac)\\ c(a^2+ab)& c(b+ab)& -abc\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}-bc& a(b+c)& a(c+b)\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|=0$

$R_1\to R_1+R_2+R_3$

$\Rightarrow \left|\begin{array}{ccc}ab+bc+ca& ab+bc+ca& ab+bc+ca\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(b+a)& -ab\end{array}\right|=0$

$\Rightarrow (ab+bc+ca)\left|\begin{array}{ccc}1.& 1& 1\\ b(a+c)& -ac& b(c+a)\\ c(a+b)& c(a+b)& -ab\end{array}\right|=0$

$c_1\to c_1-c_3,c_2\to c_2-c_3$

$\Rightarrow (ab+bc+ca)\left|\begin{array}{ccc}0.& 0& 1\\ 0& -(ac+bc+ba)& b(c+a)\\ ac+bc+ab& bc+ac+ab& -ab\end{array}\right|=0$

$\Rightarrow (ab+bc+ca)(0+(ab+bc+ca{)}^2)=0$

$\Rightarrow (ab+bc+ca{)}^3=0$

$\Rightarrow ab+bc+ca=0$