solve this problem of scalar and vectors?

Question

(AxB)2+(A.B)2=A2B2(consider only the magnitude of AxB)

This is from scalar and vector problems

Answer 1

question should be like this
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(AxB)^2+(A.B)^2=A^2 B^2
(consider only the magnitude of AxB)
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its basically a proof: showing that "for any 2 vectors A and B, the sum of the squares of their vector and scalar products" is equal to the product of magnitude of their squares.
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Left hand side
Let vector C = A cross B
|C| = |A| |B| sin (theta) >> only magnitude then
(A cross B)^2 =
(A cross B) dot (A cross B)
C dot C =
|C| |C| = |C|^2
(A cross B)^2 = |A|^2 |B|^2 sin^2 (theta) -----(1)
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(A dot B) is a scalar
= |A| |B| cos (theta), so
(A dot B)^2 = |A|^2 |B|^2 cos^2 (theta) -----(2)
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adding (1) and (2)
(AxB)^2+(A.B)^2 = |A|^2 |B|^2 {sin^2 (theta)+ cos^2 (theta)}
= |A|^2 |B|^2 {1}
= |A|^2 |B|^2 proved = RHS

Answer 2

Given two vectors A and B prove the identity.

||A X B||² + ||A • B||² = || A ||² || B ||²

First let's note some basic definitions of the cross product and the dot product. Given two vectors A and B with an angle θ between them we will define the magnitude of the cross and dot products.

Magnitude of the Cross Product of A and B
|| A X B || = || A || || B || sinθ

Magnitude of the Dot Product of A and B
|| A • B || = || A || || B || cosθ

Now let's prove the identity. We will start with the left hand side.

Left Hand Side = ||A X B||² + ||A • B||²

= (|| A || || B || sinθ)² + (|| A || || B || cosθ)²

= || A ||² || B ||² sin²θ + || A ||² || B ||² cos²θ

= (|| A ||² || B ||²) (sin²θ + cos²θ)

= (|| A ||² || B ||²) (1)

= || A ||² || B ||² = Right Hand Side

Answer 3

AXB = AB sin(AB)
A.B = AB cos (AB)

So (AxB)2+(A.B)2 = 2 AB (sin(AB) + cos (AB))

If it equals A2B2, so (sin(AB) + cos (AB)) = 2, which has no solution because if sin(AB) = 1 then cos(AB) <> 1 ...

Something I don't understand ?