Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ in the plane $x+y+z=4$.

Question

Find the image of the line $\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ with respect to plane $x+y+z=4$.

my method:

Any point in the given line will be $(3t,2t,t+1)$.

Thus the line meets the plane in $(3/2,1,3/2)$.

We now take a random point $(0,0,1)$ on this line. Clearly its image will lie on lines image wrt to this plane.

We have to find image of $(0,0,1)$.

Since the perpendicular to the plane passing through $(0,0,1)$ is $$x=y=z-1$$. It will meet the plane in $(1,1,2)$. i.e image is $(2,2,3)$.

Thus the image line which passes through $(3/2,1,3/2)$ ad $(2,2,3)$ is $$\frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}$$

---

I am however looking for alternative methods

Answer 1

An alternate method. The plane has normal vector $n=\langle 1,1,1\rangle$. The line's direction vector is $v=\langle3,2,1\rangle$. So decompose the direction vector $v$ into its two components, one parallel to $n$ and one orthogonal.

$$v_{\text{parallel to $n$}}=\frac{v\cdot n}{n\cdot n}n=\frac{6}{3}n=\langle 2,2,2\rangle$$ $$v_{\text{orthogonal to $n$}}=v-v_{\text{parallel}}=\langle 1,0,-1\rangle$$

Geometrically, the direction vector of the image line should be $v_{\text{orthogonal to $n$}}-v_{\text{parallel to $n$}}=\langle-1,-2,-3\rangle$. And as you found, the image line also passes through $(3/2,1,3/2)$. So a parametric formula for the line is $(3/2-t,1-2t,3/2-3t)$. Or in symmetric form (solving for $t$ in the three coordinates): $$\frac32-x=\frac{1-y}{2}=\frac{3/2-z}{3}$$

Answer 2

The line $r:\frac{x}{3}=\frac{y}{2}=\frac{z-1}{1}$ is the intersection of the planes $\alpha:\frac{x}{3}=\frac{y}{2};\;\beta:\frac{x}{3}=z-1$.

Any linear combination $\mathcal{P}$ of the two planes $\alpha:2x-3y=0$ and $\beta:x-3z+3=0$ passes through the line $r$ $$\mathcal{P}:h(2x-3y)+k(x-3z+3)=0;\;\forall h,k\in\mathbb{R}\land(h^2+k^2\ne 0)$$ Reorder the equation of $\mathcal{P}$ $$\mathcal{P}:x(2h+k)-3hy-3kz+3k=0$$ We have the plane passing through $r$ perpendicular to the given plane $\pi:x+y+z=4$ if the normal vectors $p=(2h+k,-3h,-3k)$ and $u=(1,1,1)$ are orthogonal, that is $p\cdot u=0$, which means $$2h+k-3h-3k=0\to h=-2k$$ Plug in the equation of $\mathcal{P}$ $$\mathcal{P*}:x(-4k+k)+6ky-3kz+3k=0\to -3k(x-2y+z-1)=0$$ The plane perpendicular to $\pi$ passing through $r$ has equation $\gamma:x-2y+z=1$.

Thus the projection $r_{\pi}$ of $r$ on $\pi$ is the line $$\begin{cases} x+y+z=4\\ x-2y+z=1\\ \end{cases} $$ that is $$r_{\pi}:\left( 3-t,1,t\right)$$ or $$ \begin{cases} x=3-z\\ y=1\\ \end{cases} $$

Answer 3

It can be solved by considering vector method also. Let the line $\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z-1}{1}$ be $A$. So the line intersects with the plane at $3/2,1,3/2$. Let us take any two points on the plane $2,1,1$ and $1,1,2$. We now form two vectors $(2-3/2,1-1,1-3/2) = (1/2,0,-1/2)$ and $(1-3/2,2-1,1-3/2) = (-1/2,1,-1/2)$.

We now find cross product of these two vectors $(1/2,0,-1/2)$ and $(-1/2,1,-1/2)$ (in order to find perpendicular line to the plane on point $3/2,1,3/2$.

The cross product is a vector whose component in
x direction is $0*(-1/2)-(-1/2)*1 = 1/2$
y direction is $(-1/2)*(-1/2)-(1/2)*(-1/2) = 1/2$
z direction is $(1/2)*1-0*(-1/2) = 1/2$
So the vector is $(1/2,1/2,1/2)$, for our convenience we take perpendicular vector as $(1,1,1)$
Take any point on line $A$, $0,0,1$ so the vector formed by point $0,0,1$ and point $3/2,1,3/2$ is $(0-3/2,0-1,1-3/2) = (-3/2,-1,-3/2)$
Now we find component of $(-3/2,-1,-3/2)$ on $(1,1,1)$. First we find magnitude of that component; that will be $\dfrac{-3/2*1+(-1)*1+(-3/2)*1}{\sqrt{1^2+1^2+1^2}} = \dfrac{-3}{\sqrt{3}}=-\sqrt{3}$,
the unit vector in the direction of $(1,1,1)$ is $(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}})$, so the component is $-\sqrt{3}(\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}},\dfrac{1}{\sqrt{3}})=(-1,-1,-1)$
So the component of vector $(-3/2,-1,-3/2)$ on plane is $(-3/2-(-1),-1-(-1),-1/2-(-1) ) = (-1/2,0,1/2)$
So the equation of the line is (using point $3/2,1,3/2$ ) is $\dfrac{x-3/2}{-1/2}=\dfrac{z-3/2}{1/2}$ and $y=1$, or $x+z=3,y=1$

Note formulae used if $A = (x_1,y_1,z_1)$ and $B=(x_2,y_2,z_2)$, two vectors, then dot product is $|A||B|cos\theta=x_1x_2+y_1y_2+z_1z_2$ so $|A|cos\theta=\dfrac{x_1x_2+y_1y_2+z_1z_2}{|B|}$, this is the magnitude of component of $A$ in the direction of $B$, to find the component we need to multiply it with the unit vector in the direction of $B$. In this case the unit vector is $\dfrac{B}{|B|}$.
Also the cross product of $A,B$ is a vector whose x component is $y_1z_2-z_1y_2$, y component is $z_1x_2-x_1z_2$, z component is $x_1y_2-y_1x_2$
Lastly if component of a vector $A$(line segment) on a line is $A_1$ then component of the vector on the plane perpendicular to the line is $A-A_1$.