# Find a unit normal to the plane containing a and b

• wolfspirit
In summary, to find a unit normal to the plane containing points a=(3;-1; 2) and b=(1; 3;-2), the equation $$\frac{a \times b}{|a \times b|}$$ can be used, where a and b are the position vectors of the given points. This will give the unit vector perpendicular to the plane.
wolfspirit

## Homework Statement

Find a unit normal to the plane containing a and b. a=(3;-1; 2) and b=(1; 3;-2)

## Homework Equations

\frac{V}{|V|} where V = a+b

## The Attempt at a Solution

(3;-1; 2) +1; 3;-2)=(4,2,0)=V
|V| = \sqrt{4^2+2^} = 2\sqrt{5}

there fore the unit vector is (\frac{2}{\sqrt{5}} \widehat{i}, \frac{1}{\sqrt{5} \widehat{j} )

but this looks like a very messy answer i am therefore not convinced i have done it right, if some one could give me some guidance that would be very much appreciated :)

ah ok i thought one could enter latex code here...

wolfspirit said:

## Homework Statement

Find a unit normal to the plane containing a and b. a=(3;-1; 2) and b=(1; 3;-2)
What plane? Two points determine a line. There are an infinite number of planes containing a given line so an infinite number of planes containing two given points.

2. Homework Equations
\frac{V}{|V|} where V = a+b

## The Attempt at a Solution

(3;-1; 2) +1; 3;-2)=(4,2,0)=V
|V| = \sqrt{4^2+2^} = 2\sqrt{5}

there fore the unit vector is (\frac{2}{\sqrt{5}} \widehat{i}, \frac{1}{\sqrt{5} \widehat{j} )

but this looks like a very messy answer i am therefore not convinced i have done it right, if some one could give me some guidance that would be very much appreciated :)
You have added the two given points. For one thing what do you mean by "adding points"? Or are you adding the position vectors of the two points? Are you given that the origin is also in this plane? But even in that case the sum of the two position vectors would be another vector in that same plane. Frankly, you seem to have completely misunderstood and misread this problem! Please reread it and then restate it here.

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wolfspirit said:
ah ok i thought one could enter latex code here...
You can- but you have to do it correctly!
1) Begin and end with [ tex ] and [ /tex ] (without the spaces) $$\int_0^\infty e^{-x^2}dx= \sqrt{\pi}$$
2) Begin and end with [ itex ] and [ /itex] in order to keep it "in line" (again without the spaces) $\int_0^\infty e^{-x^2}dx= \sqrt{\pi}$
3) Begin and end with "# #" (without the space) ##\int_0^\infty e^{-x^2}dx= \sqrt{\pi}##

HallsofIvy said:
You can- but you have to do it correctly!
1) Begin and end with [ tex ] and [ /tex ] (without the spaces)
∫∞0ex2dx=π√​
Just chipping in - you can also place a double dollar sign (i.e. \$ ) before and after the content that you want to display in latex. (It won't appear in line if you use dollar sign, just like when you use [ tex] .)

Hi,
Thanks for that!

I was using the wrong equation, i should have used $$\frac{a X b}{|aXB|}$$

wolfspirit said:
Hi,
Thanks for that!

I was using the wrong equation, i should have used $$\frac{a X b}{|aXB|}$$
So presumably the task was to find a unit normal to the plane containing a, b and the origin.

## 1. What is a unit normal to a plane?

A unit normal to a plane is a vector that is perpendicular to the plane and has a length of 1 unit. It is often denoted by n and is used to represent the orientation of the plane.

## 2. How do you find a unit normal to a plane?

To find a unit normal to a plane, you first need to determine two non-collinear vectors that lie on the plane. Let's call these vectors a and b. Then, you can use the cross product between a and b to find a vector that is perpendicular to both a and b. Finally, you can divide this vector by its magnitude to get a unit normal vector.

## 3. Why is it important to find a unit normal to a plane?

It is important to find a unit normal to a plane because it provides information about the orientation of the plane. This can be useful in various applications, such as calculating surface areas or determining angles of incidence and reflection in physics.

## 4. Can there be more than one unit normal to a plane?

No, there can only be one unit normal to a plane. This is because the unit normal is a vector that is perpendicular to the plane, and there can only be one vector that is perpendicular to a given plane.

## 5. How is a unit normal related to the equation of a plane?

The equation of a plane can be written in the form Ax + By + Cz = D, where A, B, and C represent the coefficients of the x, y, and z variables, respectively. The unit normal to this plane is given by the vector n = (A, B, C). This vector is perpendicular to the plane and can also be used to determine the orientation of the plane.

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