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 May 9th, 2017, 01:33 AM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 148 Thanks: 1 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.
 May 9th, 2017, 02:22 AM #2 Senior Member     Joined: Feb 2010 Posts: 618 Thanks: 96 C
 May 9th, 2017, 02:30 AM #3 Global Moderator   Joined: Dec 2006 Posts: 17,150 Thanks: 1282 Consider the point (1/3, 4/3, 1/3) on the sphere.
May 9th, 2017, 02:54 AM   #4
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Quote:
 Originally Posted by skipjack Consider the point (1/3, 4/3, 1/3) on the sphere.

 May 9th, 2017, 09:32 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,481 Thanks: 628 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 . 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.
 May 9th, 2017, 03:10 PM #6 Member   Joined: Feb 2015 From: Southwest Posts: 96 Thanks: 24 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.
May 9th, 2017, 09:32 PM   #7
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Quote:
 Originally Posted by Country Boy 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 . 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 ?

 May 9th, 2017, 11:00 PM #8 Global Moderator   Joined: Dec 2006 Posts: 17,150 Thanks: 1282 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.
May 10th, 2017, 01:19 AM   #9
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Quote:
 Originally Posted by skipjack 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 $D$ : $-2$

 May 10th, 2017, 03:54 AM #10 Global Moderator   Joined: Dec 2006 Posts: 17,150 Thanks: 1282 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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