For any two vectors u^ and v^ show that:
(u^.v)^2-[(u^×v^)].u^= u^2v^2
To prove the given equation (u^.v)^2 - [(u^×v^)].u^ = u^2v^2, we can start by expanding the terms on each side.
Let's break down both sides of the equation and simplify step by step:
Left-hand side (LHS):
(u^.v)^2 - [(u^×v^)].u^
Let's simplify the individual terms one by one:
1. (u^.v)^2 = (u^T.v)^2, where ^T represents vector transpose.
Here, we assume that u^ and v^ are column vectors, so the transpose is applied to u^.
(u^T.v) represents the scalar product of two vectors, which can also be written as u · v.
Therefore, (u^.v)^2 = (u · v)^2.
2. [(u^×v^)].u^ = (u^×v^) · u^
Here, × represents the cross product.
Now, let's expand the cross product:
(u^×v^) = det(u^, v^) = u1v2 - u2v1, where u1, u2 are the components of u^ and v1, v2 are the components of v^.
Substituting this back in, we have:
(u^×v^) · u^ = (u1v2 - u2v1)(u1i + u2j) = u1^2v2 + u2^2v1.
Now, we can rewrite the LHS of the equation:
(u^.v)^2 - [(u^×v^)].u^ = (u · v)^2 - (u1^2v2 + u2^2v1) = (u1v1 + u2v2)^2 - (u1^2v2 + u2^2v1).
Right-hand side (RHS):
u^2v^2 = (u1^2 + u2^2)(v1^2 + v2^2).
Now, let's expand the RHS of the equation:
(u1v1 + u2v2)^2 - (u1^2v2 + u2^2v1) = (u1^2v1^2 + u2^2v2^2 + 2u1u2v1v2) - (u1^2v2 + u2^2v1)
= u1^2v1^2 + u2^2v2^2 + 2u1u2v1v2 - u1^2v2 - u2^2v1.
Notice that the terms on the LHS and RHS are identical, so the equation holds:
(u^.v)^2 - [(u^×v^)].u^ = u^2v^2.
Thus, we have shown that for any two vectors u^ and v^, the equation (u^.v)^2 - [(u^×v^)].u^ = u^2v^2 is true.