Equation of a line perpendicular to 2 vectors

Bertrandkis
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Homework Statement



Let L be a line in R3 passing through(-1,1,2) and is perpendicular to vectors V1 (-1,1,-1)
and V2 (1,1,1). Find an equation for L in parametric form.

Homework Equations





The Attempt at a Solution


using vector V2
(x,y,z)(1,1,1)=(1,1,1)(-1,1,2)
A possible equation for L=>(x,y,z)=(-1,1,2)+t(1,1,1)

I don't know if I am right.
 
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Bertrandkis said:

Homework Statement



Let L be a line in R3 passing through(-1,1,2) and is perpendicular to vectors V1 (-1,1,-1)
and V2 (1,1,1). Find an equation for L in parametric form.

Homework Equations





The Attempt at a Solution


using vector V2
(x,y,z)(1,1,1)=(1,1,1)(-1,1,2)
A possible equation for L=>(x,y,z)=(-1,1,2)+t(1,1,1)

I don't know if I am right.
The line you give is not perpendicular to V2. Any point on (-1,1,2)+ t(1,1,1) has position vector (t-1, t+1, t+ 2) and its dot product with (1, 1, 1) is (t-1)+ (t+1)+ (t+2)= 3t+2, not 0. I'm not sure why you think that would give you a line perpendicular to (1,1,1).

But even if you were to find a line perpendicular to V2, there is no reason to think that line would also be perpendicular to V1! You need a direction vector that is perpendicular to both. Do you know how to calculate a vector perpendicular to both V1 and V2?
 
To be honest, I don't know how to find a vector perpendicular to both V1 and V2.
I can find 2 equations of a lines perpendicular to each of the vectors
(1)=> (x,y,z)=(-1,1,2)+t(-1,1,1)
(2)=> (x,y,z)=(-1,1,2)+t(1,1,1)
now where do I go from here.
 
Do you know what the cross product of two vectors is?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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