| Consider the following statements: |
| 1. R divides PQ internally in the ratio 3:2 |
| 2. R divides PQ externally in the ratio 3:2 |
| 3. Q divides PR internally in the ratio 1:2 |
| Which of the statements given above is/are correct? |
A) 1 only
B) 2 only
C) 1 and 3
D) 2 and 3
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. |
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