# Parametric equation of the orthogonal projection of a line on a plane

I have solved the following exercise but I am not sure if the method I have used to find the orthogonal projection of the line on the plane (part (b)) is correct so I would be grateful if someone could check it and tell me if it is correct. Thanks.

"(a) Write the cartesian equations of the line $$\vec{r}$$ passing through the point $$(1,0,1)$$ which intersects the $$y$$ axis and the line $$\vec{\gamma}(s)=\begin{cases}x(s)=1+s\\ y(s)=-2s\\ z(s)=3-3s\end{cases}$$;

(b) Find the orthogonal projection of the line $$\vec{r}$$ onto the plane $$8x+y+9z+1=0$$".

My solution:

(a) The plane containing all lines through the $$y$$ axis and point $$(1,0,1)$$ has normal vector $$\vec{n_{\alpha}}=(0-1,0-0,0-1)\times (0,1,0)=(1,0,-1)$$ and the plane containing all lines through $$(1,0,1)$$ and the second line has normal vector $$\vec{n_{\beta}}=(1-1,0-0,3-1)\times (1,-2,-3)=(4,2,0)$$.

So, the line through point $$(1,0,1)$$ has direction vector $$\vec{d}=\vec{n_{\alpha}}\times\vec{n_{\beta}}=(1,0,-1)\times (4,2,0)=(2,-4,2)$$ or equivalently $$(1,-2,1)$$, thus parametric equation $$\fbox{\vec{r(t)}=(1,0,1)+t(1,-2,1)}$$.

As a check, we see that $$\vec{r}(-1)=(0,2,0)$$ so $$\vec{r}$$ intersects the $$y$$ axis and also $$\vec{r}(t)=\vec{\gamma}(s)\Leftrightarrow s=t=\frac{1}{2}$$ so these two also intersect, as required.

Cartesian equations for $$\vec{r}$$ are thus $$\fbox{\begin{cases}x=z\\ x=1-\frac{y}{2}\end{cases}}$$.

(b) the projection of $$\vec{r}$$ onto the plane $$8x+y+9z+1=0$$ is given by the intersection of the plane orthogonal to the given plane and containing $$\vec{r}$$ with the given plane. A normal vector for the plane perpendicular to $$8x+y+9z+1=0$$ is $$\vec{n}=(8,1,9)\times (1,-2,1)=(19,1,-17)$$ and since it contains the line it must contain the point $$(1,0,1)$$ so an equation for it is: $$19(x-1)+1(y-0)-17(z-1)=0\Leftrightarrow 19x+y-17z-2=0$$ so the equation of the orthogonal projection of $$\vec{r}$$ is given by $$\begin{cases} 8x+y+9z=-1\\ 19x+y-17z=2\\ \end{cases}$$ $$\Leftrightarrow \fbox{\vec{r_{||}}=(\frac{3}{11},-\frac{35}{11},0)+t(\frac{26}{11},-\frac{307}{11},1)}$$

• You could perhaps save steps on part b) by, rather than projecting $\vec{r}$ onto the plane directly, projecting it onto the plane's normal (which you found earlier) and then subtracting off the resulting component. What's left must lie in the plane. Jul 11, 2021 at 8:06

The equation of the line $$r(t)$$ is correct. For the orthogonal projection, the formula for it, is

$$r'(t) = r_0 + P (r(t) - r_0))$$

Matrix P is defined as $$P = I - \dfrac{n n^T }{n^T n}$$

where $$n$$ is a normal vector to the plane of projection. In this case $$n = (8,1,9)$$

, hence $$P$$ is given by,

$$P = \begin{bmatrix} 1 && 0 && 0 \\0&&1&&0\\0&&0&&1 \end{bmatrix} - \dfrac{1}{8^2+1^2+9^2} \begin{bmatrix} 64 && 8 && 72 \\ 8 && 1 && 9\\72 && 9 && 81 \end{bmatrix} = \dfrac{1}{146} \begin{bmatrix} 82 && -8 && - 72 \\ -8 && 145 && -9 \\ -72 && - 9 && 65 \end{bmatrix}$$

Point $$r_0$$ is any point on the plane, so we can take $$r_0 = (0, -1, 0)$$, then

$$r'(t) = (0, -1, 0) + P ( (1, 0, 1) + t (1, -2, 1) - (0, -1, 0) )$$

$$r'(t) = (0, -1, 0) + P ( (1, 1, 1) + t (1, -2, 1) )$$

Simplifying by direct substitution gives

$$r'(t) = (0, -1, 0) + \dfrac{1}{73} (1, 64, -8) + t \dfrac{1}{146} ( 26, - 307, 11)$$

And finally,

$$r'(t) = \dfrac{1}{73} (1, -9, -8) + t \dfrac{1}{146} (26, -307, 11)$$

As a verification:

$$r'(t)$$ should lie in the plane $$8x + y + 9z + 1 = 0$$

Plugging $$r'(t)$$ into the quation of the plane, we get,

$$\dfrac{1}{73} (8 - 9 - 72) + 1 + t \dfrac{1}{146} (26(8)-307(1)+11(9) ) = 0$$

So indeed it does lie in the plane. The other checkpoint is $$r(t) - r'(t)$$ is along the normal to the plane.

$$r(t) - r'(t) = (1, 0, 1) - \dfrac{1}{73} (1,-9,-8) + t ( (1, -2, 1) - \dfrac{1}{146} (26, -307, 11) )$$

$$r(t) - r'(t) = \dfrac{1}{73} (72, 9, 81) + \dfrac{1}{146} t (120,15 , 135)$$

$$r(t) - r'(t) = \dfrac{9}{73} (8, 1, 9) + \dfrac{15}{146} t (8, 1, 9)$$

This shows that indeed the vector $$r(t) - r'(t)$$ is along the normal to the plane.