Saved Bookmarks
| 1. |
The symmetric equation of lines 3x+2z-5=0 and x+y-2z-3=0 is |
|
Answer» `(x-1)/(5)=(y-4)/(7)=(z-0)/(1)` `therefore 3a+2b+c=0` and a+b-2c=0 `Rightarrow (a)/(-4-1)=(b)/(1+6)=(c)/(3-2)` `Rightarrow (a)/(-5)=(b)/(7)=(c)/(1)` In order to find on the required line, we put z=0 in the two given equations to obtain 3x+2y=5 and x+y=3. On SOLVING these two equations, we get x=-1, y=4 `therefore` Coordintaes of POINT on required line are (-1,4,0) Hence, required line is `(x+1)/(-5)=(y-4)/(7)=(z-0)/(1)` |
|