1. ## geometry, parallel vectors...

QUESTION:

Show that the tangent planes of the unit sphere at the points (x,y,0) are parallel to the z axis.

MY WORK:

I get the tangent planes as x^2 + y^2 = 1 ...

in R3 that's the vector (cos(t), sin(t), z) right?

And the z axis is the vector (0,0,z)

so the dot product is z^2

And the product of the magnitudes is sqrt(1 + z^2)*z

cos(theta) = z/sqrt(1+z^2) ??

But I need cos(theta) to be 1 for those vectors to be parallel...

HELP PLEASE! What have I done?

2. Originally Posted by RanDom
QUESTION:

Show that the tangent planes of the unit sphere at the points (x,y,0) are parallel to the z axis.

MY WORK:

I get the tangent planes as x^2 + y^2 = 1 ...

in R3 that's the vector (cos(t), sin(t), z) right?

And the z axis is the vector (0,0,z)

so the dot product is z^2

And the product of the magnitudes is sqrt(1 + z^2)*z

cos(theta) = z/sqrt(1+z^2) ??

But I need cos(theta) to be 1 for those vectors to be parallel...

HELP PLEASE! What have I done?
x^2 + y^2 = 1 is definitely not the equation of a plane. I've barely glanced at this question but from my brief glance I think you'd be well advised to consider the coordinates of the points as $\displaystyle (x_1, \, y_ 1, \, 0)$ rather than (x,y,0) .......

3. OK, so...

the unit sphere is $\displaystyle x^2 + y^2 + z^2 = 1$

$\displaystyle f(x,y,z) = x^2 + y^2 + z^2 - 1$

At a point $\displaystyle P = (p_1,p_2,p_3)$ the tangent plane is

$\displaystyle 2p_1(x-p_1) + 2p_2(y-p_2) + 2p_3(z-p_3) = 0$

Cancel a factor of 2 and use $\displaystyle p_1^2 + p_2^2 + p_3^2 = 1$

$\displaystyle p_1x + p_2y + p_3 = 1$

for the point $\displaystyle (x_1,y_1,0)$ that's $\displaystyle x_1x + y_1y = 1$

... and then?

I'm really lost

4. Originally Posted by RanDom

OK, so...

the unit sphere is $\displaystyle x^2 + y^2 + z^2 = 1$

$\displaystyle f(x,y,z) = x^2 + y^2 + z^2 - 1$

At a point $\displaystyle P = (p_1,p_2,p_3)$ the tangent plane is

$\displaystyle 2p_1(x-p_1) + 2p_2(y-p_2) + 2p_3(z-p_3) = 0$

Cancel a factor of 2 and use $\displaystyle p_1^2 + p_2^2 + p_3^2 = 1$

$\displaystyle p_1x + p_2y + p_3 {\color{red}z} = 1$ Mr F adds a small correction in red.

for the point $\displaystyle (x_1,y_1,0)$ that's $\displaystyle x_1x + y_1y = 1$

... and then?

[snip]
And then you're nearly done. A normal to the plane $\displaystyle x_1x + y_1y = 1$ is $\displaystyle x_1 \, i + y_1 \, j$ ....

5. ah, which is perpendicular to (0,0,1), making the plane parallel... thanks