Question

If a, b, c are position vectors of three-collinear points such that $xa+yb+zc=0$ and atleast one scalar $x,y,z,\ne 0,$ then

• A. $x+y+z=0$
• B. $x+y+z\ne 0$
• C. there exists no relation between $x,y$ and $z$
• D. None of these

Answer 1

The correct option is A $x+y+z=0$

As $a,b,c$ are collinear.
Therefore, point $c$ divides $AB$ in a certain ration, say $y=x$
Therefore by section formula,
$c=\frac{yb+xa}{y+x}$

$\Rightarrow xa+yb=(x+y)c$

$\Rightarrow xa+yb-(x+y)c=0$
As $xa+yb+zc=0$
Therefore $x+y+z=0$