Finding the Cartesian Equation of a Plane for Vectors Exam Prep

[Anakin_k]

Homework Statement

Write the Cartesian equation for the plane containing the point (2,-1,8) and perpendicular to the line [x,y,z] = [1,-2,-3] + s[5,-4,7].

The Attempt at a Solution

The situation is that I have my Calc. + Vectors exam tomorrow morning and I'm just going through some questions on my old tests. This one has got me completely lost.

Any help to get me started would be appreciated.

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In addition, I have a second question that I thought I got correct but it was marked as incorrect so if anyone can tell me what I did wrong, it would be great.

Sketch the plan (was it a typo and not plane?) given by 3x+4y-10=0
I found the x-int to be 10/3 and y-int to be 2.5. I simply connected the two points. What did I do wrong?

 

[rock.freak667]

Homework Statement

Write the Cartesian equation for the plane containing the point (2,-1,8) and perpendicular to the line [x,y,z] = [1,-2,-3] + s[5,-4,7].

The Attempt at a Solution

The situation is that I have my Calc. + Vectors exam tomorrow morning and I'm just going through some questions on my old tests. This one has got me completely lost.

Any help to get me started would be appreciated.

If the line is perpendicular to the plane, what can you say about the direction of the line and the normal of the plane?

 

[Anakin_k]

Um I THINK they are parallel? Something in my head pops up about the normal being perpendicular to the line. So if the line is perpendicular to the plane, they are parallel?

I'm likely wrong.

 

[rock.freak667]

Um I THINK they are parallel? Something in my head pops up about the normal being perpendicular to the line. So if the line is perpendicular to the plane, they are parallel?

I'm likely wrong.

no you are right. They are parallel. If you are having trouble visualizing it, just take a sheet of paper(a plane) and two pencil, put one perpendicular to the plane (normal) and the other perpendicular to the plane will be the line.

You will see that they are both parallel.

If they are parallel, then how is the normal vector of the plane related to the direction vector the line?

 

[Anakin_k]

That makes sense.

Perhaps, they are the same? Just to be sure in the equation [x,y,z] = (-1,-2,-3) + s[5,-4,7], [5,-4,7] is the direction vector right?

If that is the case, could we use Ax+By+Cz+D=0 and plug in 5 for A; -4 for B; 7 for C? And x,y,z would be 2, -1 and 8 respectively to solve for D?

 

[rock.freak667]

That makes sense.

Perhaps, they are the same? Just to be sure in the equation [x,y,z] = (-1,-2,-3) + s[5,-4,7], [5,-4,7] is the direction vector right?

If that is the case, could we use Ax+By+Cz+D=0 and plug in 5 for A; -4 for B; 7 for C? And x,y,z would be 2, -1 and 8 respectively to solve for D?

Yes you can do that.

 

[Anakin_k]

In that case:
Ax+By+Cz+D=0
(5)(2)+(-4)(-1)+7(8)+D=0
10+4+56+D=0
D=-70

Therefore, the Cart. Eq is 5x-4y+7z-70=0. Right?

And any luck on the 2nd question?

 

[rock.freak667]

In that case:
Ax+By+Cz+D=0
(5)(2)+(-4)(-1)+7(8)+D=0
10+4+56+D=0
D=-70

Therefore, the Cart. Eq is 5x-4y+7z-70=0. Right?

And any luck on the 2nd question?

Yes that would be it.

For the second one, one you see that it can be written as 3x+4y+0z-10=0. Meaning that for any value of z, the x and y points do not depend on it. So you have the points (10/3,0,z) and (0,2.5,z); just put it number for z and join the points until you see how the plane is shaped.

 

[Anakin_k]

Oh I got it!

Gee, thanks for all the kind help Rock, I appreciate it. :)