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 }+2 j +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 }+3 j +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 |+|{ B } C |$
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.