Question

Statement I : The point $A(1,-2,-8),B(5,0,-2)$ and $C(11,3,7)$ are collinear.
Statement II : The ratio in which $B$ divides $AC$, is $2:3$

• A. Only statement I is true
• B. Only statement II is true
• C. Both statements are true
• D. Both statements are false

Answer 1

Answer: (C)

Three points $A,B$ and $C$ are collinear if $|AC|=|AB|+|BC|$
The given points are $A(1,-2,-8),B(5,0,-2)$ and $(11,3,7)$
$AB=PV$ of $B-PV$ of $A=(5\hat{i}+0\hat{j}-2\hat{k})-(\hat{i}-2\hat{j}-8\hat{k})$
$=4\hat{i}+2j+6\hat{k}$
$|AB|=\sqrt{4^2+2^2+6^2}=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}$
$BC=PV$ of $C-PV$ of $B=(11\hat{i}+3j+7\hat{k})-(5\hat{i}+0\hat{j}-2\hat{k})$
$6\hat{i}+3\hat{j}+9\hat{k}$
$|BC|=\sqrt{6^2+3^2+9^2}=\sqrt{36+9+81}=\sqrt{126}=3\sqrt{14}$
$AC=PV$ of $C-PV$ of $A=(11\hat{i}+3\hat{j}+7\hat{k})-(\hat{i}-2\hat{j}-8\hat{k})$
$=10\hat{i}+5\hat{j}+15\hat{k}$
$|AC|=\sqrt{10^2+5^2+15^2}=5\sqrt{14}$
$\therefore |AC|=|AB|+|BC|$
Thus, the given points $A,B$ and $C$ are collinear.
Let $P$ be the point $($ on the line $AC)$ which divides $AC$ in the ratio $\lambda :1$, then PV of the point
$P=\frac{\lambda \times PV\text{ of }C+1\times PV\text{ of }A}{\lambda +1}$
$=\frac{1}{\lambda +1}\{\lambda (1\hat{i}+3\hat{j}+7\hat{k})+1(\hat{i}-2\hat{j}-8\hat{k})\}$
$=\left(\frac{11\lambda +1}{\lambda +1}\right)\hat{i}+\left(\frac{3\lambda -2}{\lambda +1}\right)\hat{j}+\left(\frac{7\lambda -8}{\lambda +1}\right)\hat{k}$
$B$ lies on line $AC$ i.e., $B$ is collinear with $A$ and $C$, if $P=B$ for a unique $\lambda$
$\Rightarrow \left(\frac{11\lambda +1}{\lambda +1}\right)\hat{i}+\left(\frac{3\lambda -2}{\lambda +1}\right)\hat{j}+\left(\frac{7\lambda -8}{\lambda +1}\right)\hat{k}=5\hat{i}+0\hat{j}-2\hat{k}$
$\Rightarrow \frac{11\lambda +1}{\lambda +1}=5,\frac{3\lambda -2}{\lambda +1}=0$ and $\frac{7\lambda -8}{\lambda +1}=-2$
$\Rightarrow 11\lambda +1=5\lambda +5,3\lambda =2,7\lambda -8=-2\lambda -2$
$\Rightarrow 6\lambda =4,\lambda =\frac{2}{3},9\lambda =6$
$\Rightarrow \lambda =\frac{2}{3}$
Hence, $A,B,C$ are collinear and $B$ divides $AC$ in the ratio $\frac{2}{3}:1$ i.e. $2:3$
So, both the statements are true.