Normal Vector Equation for Plane with n=(2,-2,1) at Distance 5 from Origin

[HallsofIvy]

I would do this in a somewhat more pedestrian way. Any plane perpendicular to vector (2, -2, 1) can be written in the form 2x- 2y+ z= A. We need to find A from the condition that "the distance from (0,0,0) is 5". The distance from a plane to a point is measured along the perpendicular to the plane and the line through (0, 0, 0) perpendicular to this plane is given by x= 2t, y= -2t, z= t. Putting that into the equation of the plane, 2(2t)- 2(-2t)+ (t)= A gives 4t+ 4t+ t= 9t= A. That is, t= A/9 and so the closest point on the plane to (0,0,0) is ((2/9)A, -(2/9)A, (1/9)A). The distance from that point to (0,0,0) is $\sqrt{(4/81)A^2+ (4/81)A^2+ (1/81)A^2}= (1/3)A$. In order that that be "5" we must have A= 15. The equation of the plane is 2x- 2y+ z= 15.

In vector notation that is $(2, -2, 1)\cdot(x, y, z)- 15= 0$

 

[Physicsissuef]

What about my equation?

 

[HallsofIvy]

If you mean :

No problem. You are not wasting my time. Its the opposite, I appreciate your efforts to try to help me. This is just some confusing situation, which is not tottaly same as if we have given some point P(-2,1,2) and to find the equation, which will be
$$\frac{-2i+j+2k}{3}-3=0$$
The task of this topic, even it looks easy, its quite confusing...

It looks to me like you are subtracting a number from a vector- you can't do that! You need a dot product.

 

[Physicsissuef]

Look at the picture on the first page. p=|OP| and OP=p*$n_o$

So if $n_o=\frac{n}{|n|}$, then p should be 3.

 

[Hootenanny]

Note that usually r is the position vector of a general point such that r = xi+yj+zk. Hence, if n = ai+bj+ck then,

$$\underline{r}\cdot\underline{n} - p = 0$$

$$\left(x\underline{i} + y\underline{j} + z\underline{k}\right)\cdot\left(a\underline{i} + b\underline{j} + c\underline{k}\right)-p=0$$

$$ax+by+cz - p =0$$

Which is an equation of a plane in three space.

 

[Physicsissuef]

Note that usually r is the position vector of a general point such that r = xi+yj+zk. Hence, if n = ai+bj+ck then,

$$\underline{r}\cdot\underline{n} - p = 0$$

$$\left(x\underline{i} + y\underline{j} + z\underline{k}\right)\cdot\left(a\underline{i} + b\underline{j} + c\underline{k}\right)-p=0$$

$$ax+by+cz - p =0$$

Which is an equation of a plane in three space.

But again, something is wrong, since if p=|n|. In this case it isnt.

 

[Physicsissuef]

I would do this in a somewhat more pedestrian way. Any plane perpendicular to vector (2, -2, 1) can be written in the form 2x- 2y+ z= A. We need to find A from the condition that "the distance from (0,0,0) is 5". The distance from a plane to a point is measured along the perpendicular to the plane and the line through (0, 0, 0) perpendicular to this plane is given by x= 2t, y= -2t, z= t. Putting that into the equation of the plane, 2(2t)- 2(-2t)+ (t)= A gives 4t+ 4t+ t= 9t= A. That is, t= A/9 and so the closest point on the plane to (0,0,0) is ((2/9)A, -(2/9)A, (1/9)A). The distance from that point to (0,0,0) is $\sqrt{(4/81)A^2+ (4/81)A^2+ (1/81)A^2}= (1/3)A$. In order that that be "5" we must have A= 15. The equation of the plane is 2x- 2y+ z= 15.

In vector notation that is $(2, -2, 1)\cdot(x, y, z)- 15= 0$

H0w did u find x=2t, y=-2t, z=t?

 

[HallsofIvy]

Parametric equations of a line in the direction of vector (A, B, C) passing through point (x₀, y₀, z₀) are x= At+ x₀, y= Bt+ y₀, and z= Ct+ a₀. Didn't you learn that before the equation of a plane?

 

[Physicsissuef]

No I didn't learn it... And is n and $n_o$ are 2 different kind of vectors?

 

[Physicsissuef]

HallsofIvy?

 

[HallsofIvy]

I'm not sure what you mean by "n₀" and "n". You are referring to the "normal" vector and "unit normal". No they are not different kinds is is just that n₀ has length 1 while n can have any length.