If ^@ 3(u^2 + v^2 + w^2) = (u+v+w)^2, ^@ find the value of ^@u + v - 2w.^@
Answer:
^@0^@
- Lets try to expand and solve the equation ^@ \begin{align} & 3(u^2 + v^2 + w^2) = (u+v+w)^2 .\\ \end{align} ^@
- ^@ \begin{align} & 3(u^2 + v^2 + w^2) = (u+v+w)^2 \\ \implies & 3(u^2 + v^2 + w^2) = u^2+v^2+w^2 + 2uv + 2vw + 2wu \\ \implies & 3(u^2 + v^2 + w^2) - (u^2+v^2+w^2) = 2uv + 2vw + 2wu \\ \implies & (u^2 + v^2 + w^2)(3-1) = 2uv + 2vw + 2wu \\ \implies & 2(u^2 + v^2 + w^2) = 2(uv + vw + wu) \\ \implies & u^2 + v^2 + w^2 = uv + vw + wu \\ \implies & u^2 + v^2 + w^2 - uv - vw - wu = 0 \\ \implies & u^2 -uv + v^2 -vw + w^2 - wu = 0 \\ \implies & u(u -v) + v(v - w) + w(w - u) = 0 \\ \end{align} ^@
- As can be seen from above equation, for right hand side to be zero, there are only two possibilities. Either
(u - v) = 0, (v - w) = 0 and (w - u) = 0 ..... I
OR
u = 0, v = 0 and w = 0 ..... II - From first possibility, we can infer u = v, v = w and w = u. This implies that u = v = w .
- Second Possibility states that u = 0, v = 0, w = 0. Since this also satisfies the first possibility as well, therefore, u = v = w = 0.
- Putting the values of u, v and w to 0 in expression u + v - 2w
u + v - 2w = 0