interpretation of the equation of a plane

Recently, I found an equation **r**(s,t)= **r0** + s**v** + t**w , **

which is a plane with a point and two vectors lying on it.

I don't know why this is an equation of a plane.

can anyone teach me how to interpret r(s,t)= r0 + sv + tw ?

and

is there a method to convert** **** r(s,t)**= (-3,4,-9) +s(9,-5,9) + t(3,-2,3) into the form** (r-****r0).n**=0?

Thank you.

Re: interpretation of the equation of a plane

Do you know that there exist a unique plane given two intersecting lines? If you are given point **r**0 and vector **v**, then **r**0+ s[/b]v[/b] gives all points on the line through **r**0 in the direction of vector **v**. **r**0+ t**v** gives all points on the line through **r**0 in the direction of **w**. Those two lines, given by t= 0 and s= 0 respectively, determine the plane. Given specific values of s and v, "r0+ sv+ tw" means "starting at ro, go for a distance s in the direction of vector v, the go distance t in the direction of vector w" to get to the corresponding point on the plane.

In order that, for **r** a point in the plane, (**r**- **r**0).**n**= 0, **n** must be **perpendicular** (or **n**ormal) to the plane. And that is given by the cross product of two vectors in the plane.

In your example, (9, -5, 9) and (3, -2, 3) are vectors in the plane. Their cross product, **n**= (3, 0, -3), is perpendicular to the plane. For any point in the plane, (x, y, z), we must have ((x, y, z)- (-3, 4, -9)).(3, 0, -3)= 0 or (x+ 3)(3)+ (y+ 4)(0)+ (z+ 9)(-3)= 3x+ 9- 3z- 27= 3x- 3z- 18= 0.

Now, you were given that x= -3+ 9s+ 3t, y= 4- 5s- 2t, z= -9+ 9s+ 3t. Putting those into 3x- 3z- 18, we have 3(-3+ 9s+ 3t)+ 0(4- 5x- 2t)- 3(-9+ 9s+ 3t)- 18= (-9+ 27s+ 9t)+ (27- 27s- 9t)- 18= (-9+ 27- 18)+ (27x- 27s)+ (9t- 9t)= 0 so it does, in fact, satisfy the equation.

Re: interpretation of the equation of a plane

Quote:

Originally Posted by

**happymatthematics** Recently, I found an equation **r**(s,t)= **r0** + s**v** + t**w , **

which is a plane with a point and two vectors lying on it.

I don't know why this is an equation of a plane.

can anyone teach me how to interpret r(s,t)= r0 + sv + tw ?

and is there a method to convert** **** r(s,t)**= (-3,4,-9) +s(9,-5,9) + t(3,-2,3) into the form** (r-****r0).n**=0?

Thank you.

I have a different answer. Let $\displaystyle r = \left\langle {x,y,z} \right\rangle \;\& \,r_0 = \left\langle {x_0 ,y_0 ,z_0 } \right\rangle $

Then the plane is $\displaystyle (r-r_0)\cdot(v\times w)=0$.

Re: interpretation of the equation of a plane

Thank you HallsofIvy and Plato.

your explanations are clear and inspiring. Thank you!!