Is This the Correct Equation for a Line Perpendicular to a Plane?

[andrey21]

Find the equation of the straight line which is perpendicular to the plane.

2x - 3y +5z = 4

Which goes through the point (1,2,4)

Heres my attempt

Direction of the normal to the plane is:

(2,-3, 5)

So equation of desired line is:

r(t) = (1,2,4) + t (2,-3, 5)

Is this correct?

How do I see if (3,-1, 9) is on the line??

 

[andrey21]

Any comments would be great :)

 

[Mark44]

Find the equation of the straight line which is perpendicular to the plane.

2x - 3y +5z = 4

Which goes through the point (1,2,4)

Heres my attempt

Direction of the normal to the plane is:

(2,-3, 5)

So equation of desired line is:

r(t) = (1,2,4) + t (2,-3, 5)

Is this correct?

Yes.

How do I see if (3,-1, 9) is on the line??

If you can find a value of t for which (1, 2, 4) + t(2, -3, 5) = (3, -1, 9), then that point is on the line. If there is no such value of t, that point is not on the line.

 

[andrey21]

Fantastic I eventually arrived at a value for t=1. So the point is on the line.

 

[Mark44]

Yes, that's what I got, too.

Geometrically r(1) is a vector from the origin to the point (3, -1, 9).

 

[andrey21]

Great thanks mark 44, I do have another question if you dot mind taking a look.

I have been asked to find the fixed points of the bilinear map

f(z) = z+1/z-1

Now by setting this = z and cross multiplying I get:

z+1 = z^2 -z

Which becomes:

z^2 = 1

Now do I say the fixed points are z =1 and z=1 because it is a quadratic, or just on its own z =1 ?

 

[Mark44]

Great thanks mark 44, I do have another question if you dot mind taking a look.

I have been asked to find the fixed points of the bilinear map

f(z) = z+1/z-1

From the context below, I assume that you mean f(z) = (z + 1)/(z -1). As you wrote this, it would be interpreted as z + (1/z) - 1. When you are writing a rational expression as a single line of text, put parentheses around the whole numerator and the whole denominator if they consist of more than a single term.

Better yet, learn how to write using LaTeX.
f(z) =\frac{z+1}{z-1}
Click the equation to see my LaTeX script.

Now by setting this = z and cross multiplying I get:

z+1 = z^2 -z

Which becomes:

z^2 = 1

This equation has two solutions, one of which is not allowed here because it's not in the domain of your function f.

Now do I say the fixed points are z =1 and z=1 because it is a quadratic, or just on its own z =1 ?

See above.

 

[andrey21]

So it has only one solution??

 

[Mark44]

You have an error in your work that I didn't spot before. The equation you started with is
z = (z + 1)/(z - 1)
If you multipy both sides by z - 1, you get z² - z = z + 1, which is not equivalent to z² = 1.

There are two real fixed points for your function.

 

[andrey21]

Ah I see thank you mark 44 I didnt spot that error at all, so that would give me z^2 -2z -1 correct?

 

[Mark44]

z² - z - 1 = 0 is the equation to solve.

 

[andrey21]

Yes I ended up with z = (2+SQRT(8))/2 and z=(2-SQRT(8))/2 I substituted the back into z^2 - z -1 and they equaled zero.

 

[Mark44]

Those are the ones, but I would simplify them to
z = 1 \pm \sqrt{2}

 

[andrey21]

Ah ok so:

(2+SQRT(8))/2 becomes:

(2+SQRT(4).SQRT(2))/2

(2+2SQRT(2))/2

1+SQRT 2correct?

 

[Mark44]

Yes.