Finding plane from from the projection of a line

Question

The projection of line $$\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$$ on a plane $P$ is $$\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1}$$ Determine the equation of the plane.

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What I tried:

Clearly, $(0,1,1)$ lies on both the plane and line.

$$\cos\theta=\dfrac{2(1)+2(1)+1(-1)}{3\sqrt3}=\dfrac1{\sqrt3}\\ \sin\theta=\sqrt{\dfrac23}$$

The angle between the line and the unit normal is $\arccos\left(\sqrt{\dfrac23}\right),$ and the unit normal is perpendicular to the projected line. Using dot product:

\begin{align} (a,b,c)\cdot(2,2,1)&=3\sqrt{\dfrac23}\\ 2a+2b+c&=\sqrt6\tag1\end{align} \begin{align} (a,b,c)\cdot(1,1,-1)&=0\\ a+b-c&=0\tag2\end{align}

But they are underdetermined. What am I doing wrong?

Answer 1

When setting the condition $(a,b,c)\cdot(2,2,1) = 3\sqrt{\frac{2}{3}}$ you should also set $||(a,b,c)||=1$ , otherwise it won't give you what you want and then you'll get weird results.

But $(a,b,c),(2,2,1)$ and $(1,1,-1)$ are in the same plane, so, the cross product of $(2,2,1)$ and $(1,1,-1)$ is perpendicular to $(a,b,c)$.

$$(2,2,1) \times (1,1,-1) = (-3,3,0)$$

$$(a,b,c)\cdot(-3,3,0) = 0 \iff a=b$$

Also, as you already stated, $a+b-c=0$

$a=b$ and $a+b-c=0$ , gives $c=2a$ and $b=a$. So, $N=(a,a,2a)$. In this case, it's not necessary that $||N||$ equals $1$, so $N$ may be, for example, $(1,1,2$).

So the plane you're looking for is $x + (y-1) + 2(z-1) = 0$

Answer 2

That is a great question!

we have the two lines,

$L_1:\frac{\mathrm{x}}{2}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-1}{1}=\lambda$

Direction cosines : $(2,2,1)$ and Point on the line being : $(0,1,1)$

$L_2:\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-1}{-1}=\mu$

Direction cosines : $(1,1,-1)$ and Point on the line being : $(0,1,1)$

As $L_2$ lies on the plane, let it pass through a point 'A' whose parametric coordinates as a function of $\mu$ be $(\mu,\mu+1,-\mu+1)$, similarly Let $L_1$ pass through a point 'B' whose parametric coordinates as a function $\lambda$ be $(2\lambda,2\lambda+1,\lambda+1)$, for simplicity sake let $\lambda$ = 1 which makes B$(2,3,2)$. Now the most important fact to create any plane equation is we need a point on the plane and we need a normal vector to the plane, and luckily we do have a point on the plane as seen from $L_2$ line which passes through $(0,1,1)$ and so does the plane, now the fight is to get the normal;

Join A($\mu$) and $B(2,3,2)$ to get $AB$ vector and let us consider a situation $AB$ vector is perpendicular to the line $L_2$, hence their dot product of their direction ratios is 0, i.e.,

AB.$L_2$ =0

Substituting AB line = $(\mu-2)i+(\mu-2)j+(-\mu-1)k$ above,

On solving,$\mu$=1

How?, well the direction ratio of $L_2$ is $i+j-k$ and how was AB made? $(x_2-x_1)i+(y_2-y_1)j+(z_2-z_1)k$

So, AB vector(say n) = $-i-j-2k$, aka your normal,

The equation of plane using a point on the plane and normal to the plane is

$(r-n).a=0$ (where a = $(0,1,1)$ and $r = xi+yj+zk$)

Hence your plane equation is $x+y+2z-3=0$

Answer 3

$$(a,b,c)\cdot(2,2,1)=3\sqrt{\dfrac23}\\2a+2b+c=\sqrt6\\ (a,b,c)\cdot(1,1,-1)=0\\a+b-c=0$$

The first dot product gives $2a+2b+c=\sqrt6$, then you combine with $a+b-c=0$ to get

$$c=\frac{\sqrt6}3,~~~a+b=\frac{\sqrt6}3\tag{1}$$

Note you assume the normal vector has unit length, which means

$$a^2+b^2+c^2=1\tag{2}$$

From (1) and (2), you can solve $a$ and $b$.