Equation of a line with the given description

Question

Given a vector-valued function defined by $${\displaystyle \mathbf{r}(t)\,=\,{\begin{pmatrix}t^3+1\\t^3+1\\2t+1\end{pmatrix}}}$$

Let $\mathbb T$ denote the tangent to the curve at $A=(2,2,3)$.

Then find the equation of the line $\mathbb L$ passing through the point $u=(1,-1,2)$,parallel to the plane $2x+y+z=0$ which intersects the tangent line $\mathbb T$.

The equation of the line is in the form:

$$\mathbb L=u+\vec vs$$

Since the line is parallel to the plane,we conclude that the direction of the line is the same as the plane's,let $\vec n=(2,1,1)$ denote the normal to the plane,then $\vec n \times \vec v=(1,1,1)$,which implies:

$$(v_2-v_3,2v_3-v_1,v_1-2v_2)=1$$ So:

$$v_2-v_3=1 \tag{1}$$ $$2v_3-v_1=1$$ $$v_1-2v_2=1\tag{2}$$

moreover $\vec n \cdot \vec v=0$,which implies: $$2v_1+v_2+v_3=0 \tag{3}$$ 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 :

$$\mathbb L=(1,-1,2)+ (\frac{2}{3},-\frac{1}{6},-\frac{7}{6})s$$ With the parametric equation :

$$x=\frac{2}{3}s+1$$ $$y=-\frac{1}{6}s-1$$ $$z=-\frac{7}{6}s+2$$

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 1

I'm not sure how you got $\vec{n}\times \vec{v} = (1,1,1)$ but it is wrong.

The correct condition that $\Bbb{L} = U+t\vec{v}$ is parallel to the plane $2x+y+z=0$ is that $\vec{v}$ is orthogonal to the normal vector of the plane $\vec{n} = (2,1,1)$, or $$0 = \vec{n} \cdot \vec{v} = 2v_1+v_2+v_3.$$

The other condition is that $\mathbb{L}$ intersects the tangent $\mathbb{T}$ at the curve $\mathbf{r}$ at $A = (2,2,3) = \mathbf{r}(1)$. This tangent is given by $$\mathbb{T}=A + s\mathbf{r}'(1) = (2,2,3)+s(3,3,2), \quad s \in \Bbb{R}.$$ $\mathbb{L}$ and $\mathbb{T}$ intersect so there are $s,t \in \Bbb{R}$ such that \begin{align} U+t\vec{v} = A+s(3,3,2) &\implies \vec{AU} + t\vec{v} - s(3,3,2) = 0 \\ &\implies \vec{AU}, \vec{v}, (3,3,2) \text{ are linearly dependent} \end{align} so with $\vec{AU} = (-1,-3,-1)$ we have $$0 = \det(\vec{AU}, \vec{v}, (3,3,2)) = \begin{vmatrix} -1 & -3 & -1 \\ v_1 & v_2 & v_3 \\ 3 & 3 & 2 \end{vmatrix} = 3v_1+v_2-6v_3$$ and combining this with $2v_1+v_2+v_3 = 0$ we get $\vec{v} = (7,-15,1)$ up to multiplication by scalar. Therefore your line is given by $$\Bbb{L} = U+t\vec{v} = (1,-1,2) + t(7,-15,1), \quad t \in \Bbb{R}.$$

Answer 2

As stated, your problem can have no solution: if a line through $u=(1,-1,2)$ also passes through $A=(2,2,3)$, $\overrightarrow{Au}=(1,3,1)$ is a directing vector, and if the line is parallel to the plane with equation $2x+y+z=0$, with normal vector $\vec n=(2,1,1)$, one should have $$\overrightarrow{Au}\cdot \vec n=0,$$ which is not the case.

That's why I think the real meaning of the question is that a line through $u$, parallel to the given plane meets the tangent drawn from point $A$ on the curve.