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May 9th, 2017, 01:33 AM   #1
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Value of λ

Hello All

Can anyone help me to get the process of solving this issue:

Find the value of $\lambda$ such that the plane $2x-y+\lambda z=0$ is a tangent plane to the sphere $x^2+y^2+z^2-2x-2y-2z+2=0$ is

(A) 4 $\ $ (B) 1 $\ $ (c) 2 $\ $ (D) -2

Last edited by skipjack; May 9th, 2017 at 01:37 AM.
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May 9th, 2017, 02:22 AM   #2
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May 9th, 2017, 02:30 AM   #3
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Consider the point (1/3, 4/3, 1/3) on the sphere.
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May 9th, 2017, 02:54 AM   #4
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Quote:
Originally Posted by skipjack View Post
Consider the point (1/3, 4/3, 1/3) on the sphere.
Can you please help me to understand how did you get the point ?
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May 9th, 2017, 09:32 AM   #5
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First, Complete the squares in the equation for the sphere to write the sphere in "standard form" so you can determine the center and radius of the sphere. Second, note that the normal vector to the plane Ax+ By+ Cz= D is <A, B, C>. The line through the center of the sphere in the direction of that normal vector intersects the plane at the point where the plane is tangent to the sphere.
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May 9th, 2017, 03:10 PM   #6
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Could you use Lagrange multipliers to solve this? I feel like no because that would give you 5 unknowns and only 4 equations, right? I've been trying to solve this, and I keep hitting road blocks. I'm not sure what method should be employed.
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May 9th, 2017, 09:32 PM   #7
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Quote:
Originally Posted by Country Boy View Post
First, Complete the squares in the equation for the sphere to write the sphere in "standard form" so you can determine the center and radius of the sphere. Second, note that the normal vector to the plane Ax+ By+ Cz= D is <A, B, C>. The line through the center of the sphere in the direction of that normal vector intersects the plane at the point where the plane is tangent to the sphere.
I have written the equation in standard form and I got center $(1,1,1)$ and radius $1$. Now how do I find the normal vector ?
Please help!!!
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May 9th, 2017, 11:00 PM   #8
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A normal vector to the plane $2x - y + \lambda z = 0$ is (2, -1, $\lambda$) (or <2, -1, $\lambda$> if you prefer that notation). If you guess that $\lambda$ = 2, what is the length of that vector? You don't have to make that guess, but a lucky guess can simplify what you still need to do.
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May 10th, 2017, 01:19 AM   #9
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Quote:
Originally Posted by skipjack View Post
A normal vector to the plane $2x - y + \lambda z = 0$ is (2, -1, $\lambda$) (or <2, -1, $\lambda$> if you prefer that notation). If you guess that $\lambda$ = 2, what is the length of that vector? You don't have to make that guess, but a lucky guess can simplify what you still need to do.
I tried the following steps :

$Ax+By+Cz+D=0$

$2(1)-1+$ $\lambda$ $(1)+1=0$
$2-1+$ $\lambda$ $+1=0$
$2+$ $\lambda$ $=0$
$\lambda$ $= -2$

then the answer would be option $ D$ : $-2$

Please help me if I'm wrong in the process.
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May 10th, 2017, 03:54 AM   #10
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A line through the point (1, 1, 1) parallel to that normal vector intersects the plane at the point where the plane is tangent to the sphere. How far will that intersection point (assuming it exists) be from the point (1, 1, 1)? Hence what is the intersection point?
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