Question

If the points $A(0,3,p),B(-1,q,3),C(3,2,1)$ are collinear, then

• A. C divides AB externally in ratio $1:3$
• B. C divides AB internally in ratio $1:3$
• C. C divides AB externally in ratio $3:1$
• D. None of these

Answer 1

The correct option is D None of these

$A(0,3,p)$,$B(-1,q,3)$,$C(3,2,1)$

$C(3,2,1)$ divides $AB$ in the ratio $k:1$

$(\frac{k{x}_2+x_1}{k+1},\frac{k{y}_2+y_1}{k+1},\frac{k{z}_2+z_1}{k+1})$

$\Rightarrow C(3,2,1)=(\frac{-k+0}{k+1},\frac{kq+3}{k+1},\frac{3k+p}{k+1})$

$\Rightarrow C(3,2,1)=(\frac{-k}{k+1},\frac{kq+3}{k+1},\frac{3k+p}{k+1})$

$\Rightarrow \frac{-k}{k+1}=3$,$\frac{kq+3}{k+1}=2$,$\frac{3k+p}{k+1}=1$

$\Rightarrow -k=3k+3$,$kq+3=2k+2$,$3k+p=k+1$

$\Rightarrow -4k=3$,$kq-2k=-1$,$2k+p=1$

$\therefore k=\frac{-3}{4}$,$q-2=\frac{-1}{k}$,$2k+p=1$

$\therefore q-1=\frac{-1}{\frac{-3}{4}}$,$2\times \frac{-3}{4}+p=1$

$\therefore k=\frac{-3}{4}$,$q-1=\frac{4}{3}$,$p=1+\frac{3}{2}$

$\therefore k=\frac{-3}{4}$,$q=\frac{4}{3}+1$,$p=\frac{2+3}{2}$

Hence $k=\frac{-3}{4}$,$q=\frac{7}{3}$,$p=\frac{5}{2}$