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 → sSince 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 ) = 1So: (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 / 6So the equation of the line is : L = ( 1 , − 1 , 2 ) + ( 2 3 , − 1 6 , − 7 6 ) sWith the parametric equation : x = 2 3 s + 1 y = − 1 6 s − 1 z = − 7 6 s + 2Since 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?

Question

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      →        sSince the line is parallel to the plane,we conclude that the direction of the line is the same as the plane&#039;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    )  =  1So:                    (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      /    6So the equation of the line is :      L    =  (  1  ,  −  1  ,  2  )  +  (      2    3    ,  −      1    6    ,  −      7    6    )  sWith the parametric equation :  x  =      2    3    s  +  1  y  =  −      1    6    s  −  1  z  =  −      7    6    s  +  2Since 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&#039;t get y=2 and z=3 respectively,so where was I wrong?

Arthur Gillespie · Accepted Answer

I&#039;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    .

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.

Answered question

Answer & Explanation

New Questions in Precalculus