Given a vector-valued function defined by r(t)=((t^3+1),(t^3+1),(2t+1)) Let T denote the tangent to the curve at A=(2,2,3). Then find the equation of the line LL passing through the point u=(1,−1,2),parallel to the plane 2x+y+z=0 which intersects the tangent line TT.

smuklica8i

smuklica8i

Answered question

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

Beginner2022-07-19Added 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 .

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?