Given a vector-valued function defined by r ( t ) = ( t 3 +...

smuklica8i

smuklica8i

Answered

2022-07-18

Given a vector-valued function defined by
r ( t ) = ( t 3 + 1 t 3 + 1 2 t + 1 )
Let T denote the tangent to the curve at A = ( 2 , 2 , 3 )
Then find the equation of the line L passing through the point u=(1,−1,2),parallel to the plane 2x+y+z=0 which intersects the tangent line T
The equation of the line is in the form:
L = u + v s
Since the line is parallel to the plane,we conclude that the direction of the line is the same as the plane's,let n = ( 2 , 1 , 1 ) denote the normal to the plane,then n × v = ( 1 , 1 , 1 ), which implies:
( v 2 v 3 , 2 v 3 v 1 , v 1 2 v 2 ) = 1
So:
(1) v 2 v 3 = 1
2 v 3 v 1 = 1
(2) v 1 2 v 2 = 1
moreover n v = 0,which implies:
(3) 2 v 1 + v 2 + v 3 = 0
Substituting (1) and (2) into (3) follows:
v 1 = 2 / 3
v 2 = 1 / 6
v 3 = 7 / 6
So the equation of the line is :
L = ( 1 , 1 , 2 ) + ( 2 3 , 1 6 , 7 6 ) s
With the parametric equation :
x = 2 3 s + 1
y = 1 6 s 1
z = 7 6 s + 2
Since the line intersects the tangent line to the curve at a point with coordinate (2,2,3),we see that s=3/2,however substituting this to the y and z we don't get y=2 and z=3 respectively,so where was I wrong?

Answer & Explanation

Arthur Gillespie

Arthur Gillespie

Expert

2022-07-19Added 10 answers

I'm not sure how you got n × v = ( 1 , 1 , 1 ) but it is wrong.
The correct condition that L = U + t v is parallel to the plane 2x+y+z=0 is that v is orthogonal to the normal vector of the plane n = ( 2 , 1 , 1 ), or
0 = n v = 2 v 1 + v 2 + v 3 .
The other condition is that L intersects the tangent T at the curve r at A=(2,2,3)=r(1). This tangent is given by
T = A + s r ( 1 ) = ( 2 , 2 , 3 ) + s ( 3 , 3 , 2 ) , s R .
L and T intersect so there are s , t R such that
U + t v = A + s ( 3 , 3 , 2 ) A U + t v s ( 3 , 3 , 2 ) = 0 A U , v , ( 3 , 3 , 2 )  are linearly dependent
so with A U = ( 1 , 3 , 1 ) we have
0 = det ( A U , v , ( 3 , 3 , 2 ) ) = | 1 3 1 v 1 v 2 v 3 3 3 2 | = 3 v 1 + v 2 6 v 3
and combining this with 2 v 1 + v 2 + v 3 = 0 we get v = ( 7 , 15 , 1 ) up to multiplication by scalar. Therefore your line is given by
L = U + t v = ( 1 , 1 , 2 ) + t ( 7 , 15 , 1 ) , t R .

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