Question

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

• A. $-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a{x}_2+b{y}_2+c{z}_2+d}$
• B. $-\frac{a{x}_1\times a{x}_2+b{y}_1\times b{y}_2+c{z}_1\times c{z}_2+d^2}{a{x}_2+b{y}_2+c{z}_2+d}$
• C. $-\frac{a+b+c+d}{x_1+x_2+y_1+y_2+z_1+z_2}$
• D. $\frac{a{x}_1^2+b{y}_1^2+c{z}_1^2+d}{a{x}_2^2+b{y}_2^2+c{z}_2^2+d}$

Answer 1

The correct option is A $-\frac{a{x}_1+b{y}_1+c{z}_1+d}{a{x}_2+b{y}_2+c{z}_2+d}$

Let the plane $ax+by+cz+d=0$ divides the lines joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in the ratio $k:1$ as shown in figure.

$\therefore$Coordinates of $P(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1},\frac{k{z}_2+z_1}{k+1})$ must satisfy $ax+by+ca+d=0$

i.e., $\left(a\frac{k{x}_2+x_1}{k+1}\right)+b\left(\frac{k{y}_2+y_1}{k+1}\right)+c\left(\frac{k{z}_2+z_1}{k+1}\right)+d=0$

$\Rightarrow a(k{x}_2+x_1)+b(k{y}_2+y_1)+c(k{z}_2+z_1)+c(k{z}_2+z_1)+d(k+1)=0$

$\Rightarrow k(a{x}_2+b{y}_2+c{z}_2+d)+(a{x}_1+b{y}_1+c{z}_1+d)=0$

$\Rightarrow k=-\frac{(a{x}_1+b{y}_1+c{z}_1+d)}{(a{x}_2+b{y}_2+c{z}_2+d)}$

Hence, option A is correct.