Question

The plane $P;ax+by+cz+d=0$ divides the line joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in the ratio of $(-\frac{{ax}_1+{by}_1+{cz}_1+d}{{ax}_2+{by}_2+{cz}_2+d})=-\frac{p_1}{p_2}$; where $P_1={ax}_1+{by}_1+{cz}_1+d$ and $P_2={ax}_2+{by}_2+{cz}_2+d$. If true enter 1 else 0

Answer 1

Let the plane $ax+by+cz+d=0$ divides the line joining $(x_1,y_1,z_1)$ and$(x_2,y_2,z_2)$ in the ratio $k:1$
$\therefore$ coordinates of $p(\frac{k{x}_2+x_1}{k+1},\frac{{ky}_2+y_1}{k+1},\frac{{kz}_2+z_1}{k+1})$ must satisfy $ax+by+cz+d=0$
i.e.,$a\left(\frac{{kx}_2+x_1}{k+1}\right)+b\left(\frac{{ky}_2+y_1}{k+1}\right)+c\left(\frac{{kz}_2+z_1}{k+1}\right)+d=0$
$\Rightarrow a({kx}_2+x_1)+b({ky}_2+y_1)+c({kz}_2+z_1)+d(k+1)=0$
$\Rightarrow k({ax}_2+{by}_2+{cz}_2+d)+({ax}_1+{by}_1+{cz}_1+d)=0$

$\Rightarrow k=(-\frac{{ax}_1+{by}_1+{cz}_1+d}{{ax}_2+{by}_2+{cz}_2+d})$

$=-\frac{p_1}{p_2}$