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.
$\frac{ k ( a +2)+ a }{ k +1}= a +6$
$\Rightarrow k ( a +2)+ a =( k +1)( a +6)$
$\Rightarrow ka +2 k + a = ka +6 k + a +6$
$\Rightarrow -4 k =6$ 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: $\frac{3(b+2)-2 b}{3-2}=3 b+6-2 b=b+6$
and for z-coordinate: $\frac{3( c -2)-2 c }{3-2}=\frac{3 c -6-2 c }{1}= c - b$
Statement II is correct.
Also, let Q divide $P$ and $R$ in ratio $p : 1$ taking an
x-coordinate: $\frac{p(a+6)+a}{p+1}=a+2 \frac{p.a+6 p +a}{p+1}=a+2$
$\Rightarrow pa +6 p + a = pa + a +2 p +2$
$\Rightarrow 4 p=2 \Rightarrow p=\frac{1}{2}$
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; $\frac{(b+6) \times 1+2 b}{3}=\frac{3 b+6}{3}=b+2$
and for z-coordinate; $\frac{ c -6+2 c }{3}=\frac{3 c -6}{3}= c -2$
values are satisfied. So, statement III is correct