Find a unit normal to the plane containing a and b

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To find a unit normal to the plane containing vectors a=(3,-1,2) and b=(1,3,-2), the correct approach involves using the cross product formula, specifically $$ \frac{a \times b}{|a \times b|} $$. The initial attempt incorrectly added the vectors instead of applying the cross product. The discussion highlights confusion regarding the definition of a plane determined by two points, emphasizing that two points define a line, not a unique plane. The final clarification suggests that the task is to find a unit normal vector to the plane defined by vectors a, b, and the origin.
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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 :)
 
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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.
 
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