I have a parametric defined line as follows:


$a,b,c,d,e,f$ are constants.

otherwise stated as \begin{align} x &= a+dt \\ y &= b+et \\ z &= c+ft \end{align}

I have this parametric line, and I have this point $P(g,h,i)$ this is not on the parametric line. My question is how would you find the perpendicular plane I thought about doing it this way:

My Method:

$$A = (x,y,z)\Bigg\vert_{t=0}= (a,b,c)$$ $$B = (x,y,z)\Bigg\vert_{t=1}= (a+d,b+e,c+f)$$ $$C = (g,h,i)$$

These are 3 non-collinear points which I plan to formulate the equation of plane that is perpendicular to the parametric line. I hope since two of them are on the line that would not be an issue? Then I did the following: \begin{align} \vec{AB}&= \vec{OB}-\vec{OA} = <d,e,f> \\ \vec{BC}&= \vec{OC}-\vec{OB} = <g-(a+d), h-(b+e), i-(c+f)> \end{align} I then took the cross product of the two vectors:

$$\vec{AB} \times \vec{BC} = \det(\begin{bmatrix} \vec{i} & \vec{j} & \vec{k} \\ d & e & f \\ g-(a+d) & h-(b+e) & i-(c+f) \end{bmatrix})=\det(\begin{bmatrix} e & f \\ h-(b+e) & i-(c+f) \end{bmatrix})\vec{i}-\det(\begin{bmatrix} d & f \\ g-(a+d) & i-(c+f) \end{bmatrix})\vec{j}+\det(\begin{bmatrix} d & e \\ g-(a+d) & h-(b+e) \end{bmatrix})\vec{k}$$ which I call $\vec{n}= <m,n,o>$ With this normal vector $m, n, o$ being of course the determinants of the 2 by 2 minors of the 3 by 3 matrix. Using this normal vector I then did the following, and used this one to calculate the perpendicular plane to that line including that point P.

I began my method as the following I took the dot product of the normal vector $\vec{n}$, and $\vec{r}-\vec{r_0}$. \begin{align} \vec{n} \cdot (\vec{r}-\vec{r_0}) &= 0 \\ <m,n,o> \cdot (<x,y,z>-<g,h,i>) &= 0 \\ <m,n,o> \cdot <x-g,y-h,z-i> &= 0 \\ mx -mg +ny-nh + oz -oi &= 0 \\ mx +ny +oz -(mg +nh +oi) &= 0 \end{align}

The equation of the plane is given by the following equation: $$mx +ny+oz -(mg +nh +oi) = 0$$

Is my method correct in tackling the problem that given a line find a perpendicular plane that contains the point P?


It is correct, but you can obtain an equation of the perpendicular plane in a much simpler way:

The directing vector of the line, with coordinates $(d,e,f)$, is a normal vector of the plane. If this plane passes through the point $P(g,h,i)$, a cartesian equation is simply $$dx+ey+fz=dg+eh+fi.$$


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