Vector Algebra: Proving Mutually Perpendicular Vectors

AI Thread Summary
To prove that vectors a(v), b(v), and c(v) are mutually perpendicular, it is essential to show that the dot product a(v) · b(v) equals zero, as c(v) is already confirmed to be perpendicular to both a(v) and b(v). The relationship between the magnitudes can be established using the formula for the magnitude of the vector product from the given equations. The discussion highlights that since c(v) is the cross product of a(v) and b(v), it is inherently perpendicular to both. Ultimately, the proof confirms that b(mod) equals 1 and a(mod) equals c(mod). The vectors are thus confirmed to be mutually perpendicular.
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Homework Statement



a(v), b(v) and c(v) are three vectors. if a(v) x b(v) = c(v) and b(v) x c(v)= a(v)
Show that b(mod)= 1 and a(mod)=c(mod) and the three vectors are mutually perpendicular.
(v) denotes vector and (mod) denotes magnitude.
2. Homework Equations [/]

NA.

The Attempt at a Solution



Got some of it.Need a bit more explanation.
 
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To show that they are perpendicular, you need to show that \vec{a} \cdot \vec{b} = 0, since you already know that vector c is perpendicular to both a and b. See if you can apply this to what you are given.

As for the other two, one follows from the other. Just use the formula for magnitude of vector product on both given vector equations and compare them.

This should help:
http://en.wikipedia.org/wiki/List_of_vector_identities
 
Thanks bro.How did you know if c was perpendicular to both?
 
because C is the cross product of A and B, hence it must be perpendicular to both vectors.
 
Thanks lord.I missed such a silly thing.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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