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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)$}$

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    $\begingroup$ 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. $\endgroup$ Jul 11, 2021 at 8:06

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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.

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