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Question Number 180062 by Matica last updated on 06/Nov/22

h o w d o y o u w r i t e t h e p a r a m e t r i c e q u a t i o n o f a p l a n i n s p a c e ? i t i s g i v e n 3 p o i n t s

A (x_{A}, y_{A}, z_{A}); B (x_{B}, y_{B}, z_{B}) a n d C (x_{C}, y_{C}, z_{C}) .

Answered by hmr last updated on 07/Nov/22

Finding the parametric form of a plane

is similar to the parametric form of a line .

For line, we need an initial point

and a vector; but for plane, an initial

point and two vectors are needed .

We consider A (x_{A}, y_{A}, z_{A}) as initial point .

Then we use B & C to write vectors \vec{A B} & \vec{A C} .

\vec{A B} = < x_{B} - x_{A}, y_{B} - y_{A}, z_{B} - z_{A} >

\vec{A C} = < x_{C} - x_{A}, y_{C} - y_{A}, z_{C} - z_{A} >

Commented by hmr last updated on 07/Nov/22

Commented by hmr last updated on 07/Nov/22

Commented by hmr last updated on 07/Nov/22

Now consider another point on

the plane like D = (x, y, z) .

Notice that the vector

\vec{A D} = < x - x_{A}, y - y_{A}, z - z_{A} >

is a linear combination of

vectors \vec{A B} & \vec{A C} .

So we can write all points

of the plane like this :

\vec{A D} = t \vec{A B} + u \vec{A C}

\to

< x - x_{A}, y - y_{A}, z - z_{A} > = t < x_{B} - x_{A}, y_{B} - y_{A}, z_{B} - z_{A} > + u < x_{C} - x_{A}, y_{C} - y_{A}, z_{C} - z_{A} >

\to

{\begin{cases} x = x_{A} + t (x_{B} - x_{A}) + u (x_{C} - x_{A}) \\ y = y_{A} + t (y_{B} - y_{A}) + u (y_{C} - y_{A}) \\ z = z_{A} + t (z_{B} - z_{A}) + u (z_{C} - z_{A}) \end{cases}

Where t & u are the parameters .

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