Parametars B and D of line

[Physicsissuef]

Homework Statement

For which B and D the line $$\left\{\begin{matrix} x-2y+z-9=0 & \\ 3x+By+z+D=0 & \end{matrix}\right.$$ is in xy-plane ?

Homework Equations

The Attempt at a Solution

I found the line:

$$\frac{x-\frac{-D+9}{2}}{2-B}=\frac{y}{2}=\frac{z-\frac{D+27}{2}}{B+6}$$

What should I do next?

 

[tiny-tim]

$$\frac{x-\frac{-D+9}{2}}{2-B}=\frac{y}{2}=\frac{z-\frac{D+27}{2}}{B+6}$$

Hi Physicsissuef!

Hint: you want z to be constant, but you don't want y to be constant.

So … ?

 

[Physicsissuef]

If it is in the xy-plane probably B+6=0.

 

[tiny-tim]

But how do you stop y being infinite, then?

 

[Physicsissuef]

A point in xy plane have coordinates (x,y,0). A line in xy plane probably have $a(a_1,a_2,0)$. a is parallel vector to the line, in this case $a(2-B , 2 , B+6)$. And I don't understand you what are you talking to me...

 

[HallsofIvy]

I think you are putting far too much emphasis on plugging into formulas rather that thinking about what the problem says. In the xy plane, z is not just constant- it is specifically 0. If z= 0 then your equations become 3x- 2y- 9= 0 and 3x+ By+ D= 0. In the xy plane, those are each the equation of a line. For what B and D are those the same line?

 

[Physicsissuef]

B=-6 and D=-27. But why they need to be same?

 

[HallsofIvy]

Reread your original question! The equations represent a single line.

 

[Physicsissuef]

Ohh.... I see, thanks. And is it correct that "a" should have coordinates $a_1,a_2,0$?

 

[tiny-tim]

And I don't understand you what are you talking to me...

I was referring to the fact that your equation y/2 = (z - (D+27)/2)/(B + 6) seems to make y fixed and infinite if B + 6 = 0, while you want y to be able to take any value.

A point in xy plane have coordinates (x,y,0). A line in xy plane probably have $a(a_1,a_2,0)$. a is parallel vector to the line, in this case $a(2-B , 2 , B+6)$.

Well, yes, the line is parallel to a line through the origin with (a_1,a_2,0).

But you're concentrating too much on formulas, as HallsofIvy said.

btw, his method is a lot easier than the one you chose … if you do that in the exam, the examiner wil get the impression you're using the formula without really understanding it.

 

[Physicsissuef]

Hehe... I used the formula because I was desperate, I forgot that z=0. Thanks for the help.