Show that the particle represented by $\vec{r(t)} = \langle t^2+1, -2t+4, 4t^2+2t-3\rangle$ is contained in the plane $\pi : ax+by+cz = d$ $\forall$ $t$
I arrived at $a(t^2+1) + b(-2t+4) + c(4t^2+2t-3) = 0$ from the thought that $\vec{r(t)}$ must be orthogonal to $\vec{n}$ $\forall$ $t$ and then taking the dot product. But I can't find values for $a,b,c$ for which the expression is always $0$.
Can anyone offer any help?
On
If $\mathbf{r}(t)$ is contained in the plane $\pi$, it must be the case that $\frac{d\mathbf{r}(t)}{dt}$ (not $\mathbf{r}(t)$ as you suggested) is orthogonal to the normal vector of the plane $\mathbf{n} = (a, b , c)$:
$$ \begin{align*} \frac{d\mathbf{r}(t)}{dt} \cdot \mathbf{n} = 0 \Rightarrow a(2t) + b(-2)+c(8t+2) = 0 \end{align*} $$
By inspection, this is true when $a = -4, b = 1, c = 1$. Now we need to solve for $d$. For $t=0$, the corresponding point on the path $\mathbf{r}(t)$ is $(1,4,-3)$; plugging this into the equation for $\pi$ yields $d = -3$. Therefore, $\mathbf{r}(t)$ is contained in the plane $\pi: -4x + y + z = -3$.
you may just substitute the $(x,y,z)$ coordinates of $\vec{r(t)}$ into the equation of the plane
rewriting your equation (with $d$ included), arranged in powers of $t$ $$ (a+4c)t^2 +2(c-b)t +(a+4b-3c-d)=0 $$ this expression must be identically zero, so each coefficient vanishes. so $b=c$,$a=-4c$, $d=-3c$ and the plane is (cancelling $c$) $$ -4x+y+z=-3 $$