Question

Find all points of intersection between the vector function

$r\left(t\right)=t,t2,t3$and the plane defined by $3x-3y+z=1.$

Short answer

$3\mathrm{x}-3\mathrm{y}+\mathrm{z}=1$

$\left(1,1,1\right),\left(1,1,1\right)\mathrm{and}\left(1,1,1\right)$

Step-by-step solution

Given Information

The objective of this problem is to find all point of intersection between the vector function and the plane.

Vector function

$$\begin{gathered} \mathrm{r}\left(\mathrm{t}\right)=\left(\mathrm{t},{\mathrm{t}}^2,{\mathrm{t}}^3\right) \\ \mathrm{The}\mathrm{plane}\mathrm{is} \\ 3\mathrm{x}-3\mathrm{y}+\mathrm{z}=1 \end{gathered}$$

Calculation

Substitute $x=t,y=t^{2}andz=t^3$in the equation of plane

$$\begin{gathered} 3t-3t^2+t^3=1 \\ {\mathrm{t}}^3-1-3{\mathrm{t}}^2+3\mathrm{t}=0 \\ \left(\mathrm{t}-1\right)\left({\mathrm{t}}^2+\mathrm{t}+1\right)-3\mathrm{t}\left(\mathrm{t}-1\right)=0 \\ \left(\mathrm{t}-1\right)\left({\mathrm{t}}^2-2\mathrm{t}+1\right)=0 \\ {\left(\mathrm{t}-1\right)}^3=0 \\ \mathrm{t}=1,1,1 \end{gathered}$$

Thus the points of intersection are

localid="1650021665961" $\left(1,1,1\right),\left(1,1,1\right)\mathrm{and}\left(1,1,1\right)$

This shows that the vector function passes through the plane at a single point three times.