Show that v1 = (2/3, 1/3, 2/3) and v2 = (1/3, 2/3, -2/3) are orthonormal vectors, and find a third vector v3 for which { v1, v2, v3 } is an orthonormal set.
Two vectors u and v in Rn are said to be orthonormal if they are orthogonal
and have length 1, and a set of vectors is said to be an orthonormal set if every vector
in the set has length 1 and each pair of distinct vectors is orthogonal.
The Standard Unit Vectors in Rn Are Orthonormal
The standard unit vectors in R2 and R3 form orthonormal sets, since these vectors have length
1 and run along the coordinate axes of rectangular coordinate systems (Figure 1.2.2). More
generally, the standard unit vectors
e1 = (1, 0, 0, . . . , 0), e2 = (0, 1, 0, . . . , 0), . . . , en = (0, 0, 0, . . . , 1)
in Rn form an orthonormal set, since ei · ej = 0 if i = j and e1 = e2 = ··· = en = 1
(verify).
In the following example we form an orthonormal set of three vectors in R4.
An Orthonormal Set in R4
The vectors
q1 =1/5, 2/5, 2/5, 4/5
q2 =−2/5, 1/5,−4/5, 2/5
q3 = −4/5, 2/5, 2/5,−1/5
form an orthonormal set in R4, since
q1 = q2 = q3 = 1 and q1xq2=0, q1xq3=0, q2xq3=0
the important part is, two vectors u,v are orthogonal iff u.v = 0, and they are orthonormal iff |u| = |v| = 1.
so calculate v_{1}.v_{2}, what do you find?
what is: (2/3)*(1/3) + (1/3)*(2/3) + (2/3)*(-2/3) = ?
remember that |u| is DEFINED to be √(u.u).
so calculate v_{1}.v_{1}, and take the square root. do the same for v_{2}.
to find v_{3}, you need to solve some equations. since we want v_{3} to be orthogonal to BOTH v_{1} AND v_{2}, this means:
v_{1}.v_{3} = 0
v_{2}.v_{3} = 0.
if v_{3} = (x,y,z) this translates to the two equations:
(2/3)x + (1/3)y + (2/3)z = 0
(1/3)x + (2/3)y + (-2/3)z = 0
solve this system for x,y and z. you should get an answer that depends on one free parameter. then use that |v_{3}| = 1 to determine what that parameter should be.
while v_{1}xv_{2} is orthogonal to both v_{1} and v_{2}, this approach only works in 3 dimensions (3 is special).
the reason it works is because:
[a.(axb)]a = (axa) x (axb) = 0x(axb) = 0, forcing a.(axb) = 0, if a is non-zero. and
[b.(bxa)]b = (bxb) x (bxa) = 0x(bxa) = 0, forcing b.(bxa) = 0, so that b.(axb) = -b.(bxa) = -0 = 0.
however, orthogonality works in more (or less) dimensions than just three. not all vectors represent "physical" things, these are MATHEMATICAL concepts, and reducing them to "the real world" reduces their power.
Deveno,
You wrote: "while v1Xv2 is orthogonal to both v1 and v2....."
If v1Xv2 is orthogonal to v1 and v2, then obviously v1.(v1Xv2) = 0 and v2.(v1Xv2) = 0.
(if a is perpendicular b then a.b =0 by definmition of orthogonality)
What is the point of using some very complicated formula? I doubt very many people would recognize it or recall it in the incomplete form you gave, or even in the complete form for that matter. I don't see what insight this gives.
What "very complicated formula" are you talking about? Deveno did not give a formula for calculating anything, nor did he start from the assumption that "v1Xv2 is orthogonal to v1 and v2", He was showing why the cross product of v1 and v2 is orthogonal to both v1 and v2.
(axb)x(cxd) = (a.bxc)b – (a.bxc)d
To prove this formula without recourse to geometry (pictures), you have to start with the definition
aXb = (a2b3-b2a3, b1a3-a1b3, a1b2-a2b1)
and then do a massive amount of algebra.
On the other hand
a.(aXb) = 0 by taking dot product with definition.
And anyhow, how do you know aXa = 0? You have to start from the definiton of aXb above, and then it’s just as easy to do a.(aXb).
EDIT: If you're going to discuss aXb, it's a good idea to start with the definition. It saves a lot of confusion.
EDIT: Incidentally, from the definition of aXb above it is apparent that a.bXc is det(a,b,c) which would also immediateley show that a.bxa is zero because two rows of the determinant are equal. I hesitate to use it because it is not transparent.
Show that v1 and v2 are orthonormal:
for this just show that v1 . v2=0 and that the magnitude of v1=magnitude of v2=1
For the last part:
create a system of linear equations such that: v1 . v3=0, v2 . v3=0 and the magnitude of v3=1
the magnitude of a vector x=(x1^2+x2^2+x3^2)^(1/2)
Hope this helps
peruvian, that was covered by Deveno. My point was that v3 = v1Xv2 is easier if you are familiar with cross product.
If the problem was posed as an excercise in linear algebra, Deveno is correct, as an excercise in vector analysis, the cross product is easier.