smuklica8i

Answered

2022-07-18

Given a vector-valued function defined by

$\mathbf{r}(t)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left(\begin{array}{c}{t}^{3}+1\\ {t}^{3}+1\\ 2t+1\end{array}\right)$

Let $\mathbb{T}$ 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 $\mathbb{T}$

The equation of the line is in the form:

$\mathbb{L}=u+\overrightarrow{v}s$

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

$({v}_{2}-{v}_{3},2{v}_{3}-{v}_{1},{v}_{1}-2{v}_{2})=1$

So:

$\begin{array}{}\text{(1)}& {v}_{2}-{v}_{3}=1\end{array}$

$2{v}_{3}-{v}_{1}=1$

$\begin{array}{}\text{(2)}& {v}_{1}-2{v}_{2}=1\end{array}$

moreover $\overrightarrow{n}\cdot \overrightarrow{v}=0$,which implies:

$\begin{array}{}\text{(3)}& 2{v}_{1}+{v}_{2}+{v}_{3}=0\end{array}$

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?

$\mathbf{r}(t)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left(\begin{array}{c}{t}^{3}+1\\ {t}^{3}+1\\ 2t+1\end{array}\right)$

Let $\mathbb{T}$ 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 $\mathbb{T}$

The equation of the line is in the form:

$\mathbb{L}=u+\overrightarrow{v}s$

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

$({v}_{2}-{v}_{3},2{v}_{3}-{v}_{1},{v}_{1}-2{v}_{2})=1$

So:

$\begin{array}{}\text{(1)}& {v}_{2}-{v}_{3}=1\end{array}$

$2{v}_{3}-{v}_{1}=1$

$\begin{array}{}\text{(2)}& {v}_{1}-2{v}_{2}=1\end{array}$

moreover $\overrightarrow{n}\cdot \overrightarrow{v}=0$,which implies:

$\begin{array}{}\text{(3)}& 2{v}_{1}+{v}_{2}+{v}_{3}=0\end{array}$

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 & Explanation

Arthur Gillespie

Expert

2022-07-19Added 10 answers

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

The correct condition that $\mathbb{L}=U+t\overrightarrow{v}$ is parallel to the plane 2x+y+z=0 is that $\overrightarrow{v}$ is orthogonal to the normal vector of the plane $\overrightarrow{n}=(2,1,1)$, or

$0=\overrightarrow{n}\cdot \overrightarrow{v}=2{v}_{1}+{v}_{2}+{v}_{3}.$

The other condition is that $\mathbb{L}$ intersects the tangent $\mathbb{T}$ at the curve r at A=(2,2,3)=r(1). This tangent is given by

$\mathbb{T}=A+s{\mathbf{r}}^{\prime}(1)=(2,2,3)+s(3,3,2),\phantom{\rule{1em}{0ex}}s\in \mathbb{R}.$

$\mathbb{L}$ and $\mathbb{T}$ intersect so there are $s,t\in \mathbb{R}$ such that

$\begin{array}{rl}U+t\overrightarrow{v}=A+s(3,3,2)& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overrightarrow{AU}+t\overrightarrow{v}-s(3,3,2)=0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overrightarrow{AU},\overrightarrow{v},(3,3,2)\text{are linearly dependent}\end{array}$

so with $\overrightarrow{AU}=(-1,-3,-1)$ we have

$0=det(\overrightarrow{AU},\overrightarrow{v},(3,3,2))=\left|\begin{array}{ccc}-1& -3& -1\\ {v}_{1}& {v}_{2}& {v}_{3}\\ 3& 3& 2\end{array}\right|=3{v}_{1}+{v}_{2}-6{v}_{3}$

and combining this with $2{v}_{1}+{v}_{2}+{v}_{3}=0$ we get $\overrightarrow{v}=(7,-15,1)$ up to multiplication by scalar. Therefore your line is given by

$\mathbb{L}=U+t\overrightarrow{v}=(1,-1,2)+t(7,-15,1),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}.$

The correct condition that $\mathbb{L}=U+t\overrightarrow{v}$ is parallel to the plane 2x+y+z=0 is that $\overrightarrow{v}$ is orthogonal to the normal vector of the plane $\overrightarrow{n}=(2,1,1)$, or

$0=\overrightarrow{n}\cdot \overrightarrow{v}=2{v}_{1}+{v}_{2}+{v}_{3}.$

The other condition is that $\mathbb{L}$ intersects the tangent $\mathbb{T}$ at the curve r at A=(2,2,3)=r(1). This tangent is given by

$\mathbb{T}=A+s{\mathbf{r}}^{\prime}(1)=(2,2,3)+s(3,3,2),\phantom{\rule{1em}{0ex}}s\in \mathbb{R}.$

$\mathbb{L}$ and $\mathbb{T}$ intersect so there are $s,t\in \mathbb{R}$ such that

$\begin{array}{rl}U+t\overrightarrow{v}=A+s(3,3,2)& \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overrightarrow{AU}+t\overrightarrow{v}-s(3,3,2)=0\\ & \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\overrightarrow{AU},\overrightarrow{v},(3,3,2)\text{are linearly dependent}\end{array}$

so with $\overrightarrow{AU}=(-1,-3,-1)$ we have

$0=det(\overrightarrow{AU},\overrightarrow{v},(3,3,2))=\left|\begin{array}{ccc}-1& -3& -1\\ {v}_{1}& {v}_{2}& {v}_{3}\\ 3& 3& 2\end{array}\right|=3{v}_{1}+{v}_{2}-6{v}_{3}$

and combining this with $2{v}_{1}+{v}_{2}+{v}_{3}=0$ we get $\overrightarrow{v}=(7,-15,1)$ up to multiplication by scalar. Therefore your line is given by

$\mathbb{L}=U+t\overrightarrow{v}=(1,-1,2)+t(7,-15,1),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}.$

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