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

Answer 1

Finding the condition:

Given $a,b,c$ are position vectors of three collinear points ,therefore three points lie in a straight line

Say point $c$ divides the line in a ratio$y:x$

Therefore, using section formula,

$c=\frac{(yb+xa)}{(y+x)}$

$c(x+y)=(yb+xa)$or we can write

$xa+yb-(x+y)c=0$

Since $xa+yb+zc=0$

Therefore $x+y+z=0$

Hence, the condition is $\mathit{x}\mathbf{+}\mathit{y}\mathbf{+}\mathit{z}\mathbf{=}\mathbf{0}\mathbf{.}$