Question

The plane $ax+by+cz+d=0$ divides the line joining the points $(x_1,y_1,z_1)$ 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. $\frac{(a{x}_1+b{y}_1+c{z}_1+d)}{(a{x}_2+b{y}_2+c{z}_2+d)}$
• C. $\frac{a{x}_1{x}_2+b{y}_1{y}_2+c{z}_1{z}_2}{d^2}$
• D. None of these

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 given plane meet the line joining the given points in $(x_3,y_3,z_3)$.
Then $a{x}_3+b{y}_3+c{z}_3+d=0$ ...(1)
Also let the point $(x_3,y_3,z_3)$ divide the line joining the given points in the ratio $m:n$
Then

$x_3=\frac{m{x}_1+n{x}_2}{m+n},y_3=\frac{m{y}_1+n{y}_2}{m+n}$ and $z_3=\frac{m{z}_1+n{z}_2}{m+n}$
Substituting these values in (1), we get
$a\left(\frac{m{x}_1+n{x}_2}{m+n}\right)+b\left(\frac{m{y}_1+n{y}_2}{m+n}\right)+c\left(\frac{m{z}_1+n{z}_2}{m+n}\right)+d=0$
$\Rightarrow a(m{x}_1+n{x}_2)+b(m{y}_1+n{y}_2)+c(m{z}_1+n{z}_2)+d(m+n)=0$

$\Rightarrow m(a{x}_1+b{y}_1+c{z}_1+d)+n(a{x}_2+b{y}_2+c{z}_2+d)=0$
$\Rightarrow \frac{n}{m}=\frac{-(a{x}_1+b{y}_1+c{z}_1+d)}{(a{x}_2+b{y}_2+c{z}_2+d)}$

Hence, option A.