Three Dimensional Geometry Question 330
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:
- R divides PQ internally in the ratio 3:2
- R divides PQ externally in the ratio 3:2
- Q divides PR internally in the ratio 1:2 Which of the statements given above is/are correct?
Options:
A) 1 only
B) 2 only
C) 1 and 3
D) 2 and 3
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Given that P (a, b, c), Q(a+2,b+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 is divided P and Q externally in 3:2 ratio. Putting this value for y- and z-coordinated 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 (2) is correct. Also, let Q divide P and R in ratio P : 1 taking an x-co-ordinate: $ \frac{p(a+6)+a}{p+1}=a+2 $
$ \Rightarrow \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 but the division is internal and in the ratio 1 : 2 Verifying for y-and z-coordinates, satisfies this results. For y co-ordinate, $ \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 (3) is correct.
BETA
