Given the surface #f(x,y,z)=y^2 + 3 x^2 + z^2 - 4=0# and the points #p_1=(2,1,1)# and #p_2=(3,0,1)# determine the tangent plane to #f(x,y,z)=0# containing the points #p_1# and #p_2#?

Answer 1

See below.

Calling
#Sigma->f(x,y,z)=y^2 + 3 x^2 + z^2 - 4=0#
and considering #p = (x,y,z)# such that
#p in Sigma#, we have

#vec n = (p-p_1) xx (p-p_2)# is a vector normal to the plane
#Pi# defined by the points #p, p_1, p_2#

Now, the vector #vec n# can be computed over #Sigma# as #grad f = ((partial f)/(partial x), (partial f)/(partial y), (partial f)/(partial z)) =2(3x,y,z) #
The #Sigma# tangent point is then determined by the equations

#{ (n_x/norm(vec n)=pmf_x/norm(grad f)), (<< grad f, p_1-p_2 >> = 0), (f(x,y,z)=0) :}#(1)

or

#{((1 - z)/sqrt[(x + y-3)^2 + 2 (z-1)^2] pm (3 x)/sqrt[ 9 x^2 + y^2 + z^2]=0),(y-3x=0),(y^2 + 3 x^2 + z^2 - 4=0):}#(2)

Solving for #x,y,z# we obtain

#p_(t_1) = (0.531359, 1.59408, -0.782233)# and
#p_(t_2) = (0.242834, 0.728503, 1.81449)# the tangency points

and also

#(grad _1 f)/norm(grad_1 f) =(vec n_1)/norm(vec n_1) = (0.668034, 0.668034, -0.327812) #
#(grad_2 f)/norm(grad_2 f) =(vec n_2)/norm(vec n_2)=-(0.349138, 0.349138, 0.869601)#

the corresponding normal surface vectors.

Note:

a) In (1,2) we consider only a vector component. The choice is one of

#(((1 - z)/sqrt[(x + y-3)^2 + 2 (z-1)^2] - (3 x)/sqrt[ 9 x^2 + y^2 + z^2]),((1 - z)/sqrt[(x + y-3)^2 + 2 (z-1)^2] - y/sqrt[ 9 x^2 + y^2 + z^2]),((x + y-3)/sqrt[(x + y-3)^2 + 2 (z-1)^2] - z/sqrt[ 9 x^2 + y^2 + z^2]))#

b) The sign #pm# before the gradient is to qualify the two solutions to the problem.

c) Here #(cdot xx cdot)# represents the vector product and #<< cdot, cdot >># the scalar product. Also #norm(cdot)# represents the vector norm.

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Answer 2

To determine the tangent plane to the surface ( f(x,y,z) = y^2 + 3x^2 + z^2 - 4 = 0 ) containing the points ( p_1 = (2,1,1) ) and ( p_2 = (3,0,1) ), we need to follow these steps:

  1. Find the gradient vector of the surface function ( f(x,y,z) ).
  2. Evaluate the gradient vector at the points ( p_1 ) and ( p_2 ) to obtain the normal vectors to the surface at these points.
  3. Use the normal vectors and the points ( p_1 ) and ( p_2 ) to find the equations of the tangent planes at these points.
  4. Equate the equations of the tangent planes to find the common tangent plane passing through both ( p_1 ) and ( p_2 ).

First, let's find the gradient vector of ( f(x,y,z) ):

[ \nabla f(x,y,z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) ]

[ \nabla f(x,y,z) = \left( 6x, 2y, 2z \right) ]

Evaluate the gradient vector at ( p_1 ) and ( p_2 ):

At ( p_1 = (2,1,1) ):

[ \nabla f(2,1,1) = \left( 12, 2, 2 \right) ]

At ( p_2 = (3,0,1) ):

[ \nabla f(3,0,1) = \left( 18, 0, 2 \right) ]

The normal vector to the surface at ( p_1 ) is ( \left( 12, 2, 2 \right) ), and at ( p_2 ) is ( \left( 18, 0, 2 \right) ).

Now, we can use the point-normal form of the equation of a plane to find the equations of the tangent planes at ( p_1 ) and ( p_2 ):

At ( p_1 = (2,1,1) ), the equation of the tangent plane is:

[ 12(x - 2) + 2(y - 1) + 2(z - 1) = 0 ]

At ( p_2 = (3,0,1) ), the equation of the tangent plane is:

[ 18(x - 3) + 0(y - 0) + 2(z - 1) = 0 ]

Now, equate the equations of the tangent planes to find the common tangent plane passing through both ( p_1 ) and ( p_2 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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