Question

If `veca` and `vecb` are two vectors such that `|veca + vecb| = |vecb|`, then prove that `(veca + 2vecb)` is perpendicular to `veca`.

Answer 1

Given, `|veca + vecb| = |vecb|`

On squaring both sides, we get

`|veca + vecb|^2 = |vecb|^2`

⇒ `|veca|^2 + |vecb|^2 + 2|veca||vecb| = |vecb|^2`

⇒ `|veca|^2 + 2|veca||vecb|` = 0

⇒ `|veca|.(|veca| + 2|vecb|)` = 0

⇒ `veca.(veca + 2vecb)` = 0

Since, dot product of `veca` and `veca + 2vecb` is zero, thus vectors are perpendicular.

Hence proved