I have got this line g : x → = ( − 1 2 0 ) + s ⋅ ( − 1 4 − 1 ) And this plane: E t : t x + y + t z = 0Now for which t is the line element of the plane?My calculations: When I put the equation of the line in the equation of the plane i get: And this plane: t ( − 1 − s ) + ( 2 + 4 s ) + t ( − 2 ) = 0This is the same as: 2 − t + s ( 4 − 2 t ) = 0 =&gt; s = t − 2 4 − 2 t = − 1 2 And now I do not know how to go on.The solution is t=2Can someone explain the last step to the solution please?

Question

I have got this line      g    :                      x                    →      =        (                            −          1                                      2                                      0                      )      +        s    ⋅      (                            −          1                                      4                                      −          1                      )  And this plane:      E    t    :      t  x  +  y  +  t  z  =  0Now for which t is the line element of the plane?My calculations: When I put the equation of the line in the equation of the plane i get: And this plane:  t  (  −  1  −  s  )  +  (  2  +  4  s  )  +  t  (  −  2  )  =  0This is the same as:  2  −  t  +  s  (  4  −  2  t  )  =  0  =&amp;gt;  s  =                    t        −        2                    4        −        2        t              =  −            1      2      And now I do not know how to go on.The solution is t=2Can someone explain the last step to the solution please?

kartonaun · Accepted Answer

The line and the plane are parallel iff the plane&#039;s normal vector (t,1,t) and the line&#039;s direction vector (−1,4,−1) are perpendicular; this gives t=2.As for t=2 the vector (−1,2,0) lies in the plane the whole line lies in that plane.

Taniya Burns · Accepted Answer

Calling  p  =  (  x  ,  y  ,  z      )          †                    n      →        =  (  t  ,  1  ,  t      )          †              p    0    =  (  −  1  ,  2  ,  0      )          †                    v      →        =  (  −  1  ,  4  ,  −  1      )          †      we have the plane   Π  →  p  ⋅            n      →        =  0 and the line   L  →  p  =      p    0    +  s            v      →       . Now if   L  ∈  Π we have  (      p    0    +  s            v      →        )  ⋅            n      →        =  0  ⇒      {                                        p            0                    ⋅                                    n              →                                =          0                                                                v              →                                ⋅                                    n              →                                =          0                        From the second equation we have   −  t  +  4  −  t  =  0  ⇒  t  =  2 and now making       p    0    ⋅            n      →        =  2  ×  (  −  1  )  +  1  ×  2  +  2  ×  0  =  0 hence for   t  =  2  ⇒  L  ∈  Π

What Plane contains the line? g: vec(x)=((-1),(2),(0))+s * ((-1),(4),(-1))

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