smuklica8i

2022-07-18

Given a vector-valued function defined by
$\mathbf{r}\left(t\right)\phantom{\rule{thinmathspace}{0ex}}=\phantom{\rule{thinmathspace}{0ex}}\left(\begin{array}{c}{t}^{3}+1\\ {t}^{3}+1\\ 2t+1\end{array}\right)$
Let $\mathbb{T}$ denote the tangent to the curve at $A=\left(2,2,3\right)$
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:
$\mathbb{L}=u+\stackrel{\to }{v}s$
Since the line is parallel to the plane,we conclude that the direction of the line is the same as the plane's,let $\stackrel{\to }{n}=\left(2,1,1\right)$ denote the normal to the plane,then $\stackrel{\to }{n}×\stackrel{\to }{v}=\left(1,1,1\right)$, which implies:
$\left({v}_{2}-{v}_{3},2{v}_{3}-{v}_{1},{v}_{1}-2{v}_{2}\right)=1$
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 $\stackrel{\to }{n}\cdot \stackrel{\to }{v}=0$,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 :
$\mathbb{L}=\left(1,-1,2\right)+\left(\frac{2}{3},-\frac{1}{6},-\frac{7}{6}\right)s$
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?

Arthur Gillespie

Expert

I'm not sure how you got $\stackrel{\to }{n}×\stackrel{\to }{v}=\left(1,1,1\right)$ but it is wrong.
The correct condition that $\mathbb{L}=U+t\stackrel{\to }{v}$ is parallel to the plane 2x+y+z=0 is that $\stackrel{\to }{v}$ is orthogonal to the normal vector of the plane $\stackrel{\to }{n}=\left(2,1,1\right)$, or
$0=\stackrel{\to }{n}\cdot \stackrel{\to }{v}=2{v}_{1}+{v}_{2}+{v}_{3}.$
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
$\mathbb{T}=A+s{\mathbf{r}}^{\prime }\left(1\right)=\left(2,2,3\right)+s\left(3,3,2\right),\phantom{\rule{1em}{0ex}}s\in \mathbb{R}.$
$\mathbb{L}$ and $\mathbb{T}$ intersect so there are $s,t\in \mathbb{R}$ such that

so with $\stackrel{\to }{AU}=\left(-1,-3,-1\right)$ we have
$0=det\left(\stackrel{\to }{AU},\stackrel{\to }{v},\left(3,3,2\right)\right)=|\begin{array}{ccc}-1& -3& -1\\ {v}_{1}& {v}_{2}& {v}_{3}\\ 3& 3& 2\end{array}|=3{v}_{1}+{v}_{2}-6{v}_{3}$
and combining this with $2{v}_{1}+{v}_{2}+{v}_{3}=0$ we get $\stackrel{\to }{v}=\left(7,-15,1\right)$ up to multiplication by scalar. Therefore your line is given by
$\mathbb{L}=U+t\stackrel{\to }{v}=\left(1,-1,2\right)+t\left(7,-15,1\right),\phantom{\rule{1em}{0ex}}t\in \mathbb{R}.$

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