Question

$P(a,b,c);Q(a+2,b+2,c-2)$ and $R(a+6,b+6,c-6)$ are
collinear. Consider the following statements:
I. R divides PQ internally in the ratio $3:2$
II. R divides PQ externally in the ratio $3:2$
III. Q divides PR internally in the ratio $1:2$
Which of the statements given above is/are correct?

• A. Only I
• B. Only II
• C. I and III
• D. II and III

Answer 1

Answer: (D)

Given that $P(a,b,c),Q(a+2,b+2,c-2)$ and $R(a+6,b+6,c-6)$ are collinear, one point must divide, the other two points externally or internally. Let $R$ divide $P$ and $Q$ in ratio $k:1$ so, taking on x-coordinates.
$\frac{k(a+2)+a}{k+1}=a+6$
$\Rightarrow k(a+2)+a=(k+1)(a+6)$
$\Rightarrow ka+2k+a=ka+6k+a+6$
$\Rightarrow -4k=6$ or $k=-\frac{3}{2}$
Negative sign shows that this is external division in ratio $3:2$. So, $R$ divides $P$ and $Q$ externally in $3:2$ ratio. Putting this value for $y$ - and $z$ - coordinates satisfied :
for $y$ - coordinate: $\frac{3(b+2)-2b}{3-2}=3b+6-2b=b+6$
and for z-coordinate: $\frac{3(c-2)-2c}{3-2}=\frac{3c-6-2c}{1}=c-b$
Statement II is correct.
Also, let Q divide $P$ and $R$ in ratio $p:1$ taking an
x-coordinate: $\frac{p(a+6)+a}{p+1}=a+2\frac{p.a+6p+a}{p+1}=a+2$
$\Rightarrow pa+6p+a=pa+a+2p+2$
$\Rightarrow 4p=2\Rightarrow p=\frac{1}{2}$
Positive sign shows that the division is internal and in the ratio $1:2$
Verifying for $y$ - and $z$ - coordinates, satisfies this results.
For y coordinate; $\frac{(b+6)\times 1+2b}{3}=\frac{3b+6}{3}=b+2$
and for z-coordinate; $\frac{c-6+2c}{3}=\frac{3c-6}{3}=c-2$
values are satisfied. So, statement III is correct