Linear algebra vector problem

[math1818]

Hello!
I'm unable to sovle the question. I don't know what approach to take.

The question: two vectors u and v has the coordinates (1,2,3 ) u . v (-1,-1, 1). Determine a unit vector that is ortogonal to the both U and V.
I sinicerly thank you for the help.

 

[Deleted member 4993]

Hello!
I'm unable to sovle the question. I don't know what approach to take.

The question: two vectors u and v has the coordinates (1,2,3 ) u . v (-1,-1, 1). Determine a unit vector that is ortogonal to the both U and V.
I sinicerly thank you for the help.

Have you thought about using cross-product of vectors?

 

[pka]

The question: two vectors u and v has the coordinates (1,2,3 ) u . v (-1,-1, 1). Determine a unit vector that is ortogonal to the both U and V.

Given vectors \(\vec{U}=<1,2,3>~\&~\vec{V}=<1,1,1>\). Then \(\vec{U}\times\vec{V}\) perpendicular to both. SEE HERE
If \(\vec{A}\bot\vec{B}\) then \(-\vec{A}\bot\vec{B}\). Given a vector \(\vec{C}\ne\vec{O}\) then \(\dfrac{\vec{C}}{\|\vec{C}\|}\) is a unit vector(normalized).

 

[HallsofIvy]

Another way to do this is to write your desired vector as <x, y, z>. The requirement that it be orthogonal to <1, 2, 3> is that x+ 2y+ 3z= 0 and the requirement that it be orthogonal to <1, 1, 1> is that x+ y+ z= 0.

That is, of course, two equations in three unknowns. There are an infinite number of such vectors. The requirement that it be a unit vector adds the equation \(\displaystyle x^2+ y^2+ z^2= 1\).
If we subtract x+ y+ z= 0 from x+ 2y+ 3z= 0 we eliminate x and get y+ 2z= 0 so y= -2z.

Then x+ y+ z= x- 2z+ z= x- z= 0 so x= z.

Finally, \(\displaystyle x^2+ y^2+z^2= z^2+ 4z^2+ z^2= 6z^2= 1\).

Notice that there will be two solutions for z. There will be two unit vectors that are orthogonal to <1, 2, 3> and <1, 1, 1> in opposite directions.