Find $$3-$$dimensional vectors $${\overrightarrow v _1},{\overrightarrow v _2},{\overrightarrow v _3}$$ satisfying
$$\,{\overrightarrow v _1}.{\overrightarrow v _1} = 4,\,{\overrightarrow v _1}.{\overrightarrow v _2} = - 2,\,{\overrightarrow v _1}.{\overrightarrow v _3} = 6,\,\,{\overrightarrow v _2}.{\overrightarrow v _2}$$
$$ = 2,\,{\overrightarrow v _2}.{\overrightarrow v _3} = - 5,\,{\overrightarrow v _3}.{\overrightarrow v _3} = 29$$

Let $$\overrightarrow A \left( t \right) = {f_1}\left( t \right)\widehat i + {f_2}\left( t \right)\widehat j$$ and $$$\overrightarrow B \left( t \right) = {g_1}\left( t \right)\overrightarrow i + {g_2}\left( t \right)\widehat j,t \in \left[ {0,1} \right],$$$
where $${f_1},{f_2},{g_1},{g_2}$$ are continuous functions. If $$\overrightarrow A \left( t \right)$$ and $$\overrightarrow B \left( t \right)$$ are nonzero vectors for all $$t$$ and $$\overrightarrow A \left( 0 \right) = 2\widehat i + 3\widehat j,$$ $$\,\overrightarrow A \left( 1 \right) = 6\widehat i + 2\widehat j,$$ $$\,\overrightarrow B \left( 0 \right) = 3\widehat i + 2\widehat j$$ and $$\,\overrightarrow B \left( 1 \right) = 2\widehat i + 6\widehat j.$$ Then show that $$\,\overrightarrow A \left( t \right)$$ and $$\,\overrightarrow B \left( t \right)$$ are parallel for some $$t.$$

Let $$u$$ and $$v$$ be units vectors. If $$w$$ is a vector such that $$w + \left( {w \times u} \right) = v,$$ then prove that $$\left| {\left( {u \times v} \right) \cdot w} \right| \le 1/2$$ and that the equality holds if and only if $$u$$ is perpendicular to $$v .$$

For any two vectors $$u$$ and $$v,$$ prove that
(a) $${\left( {u\,.\,v} \right)^2} + {\left| {u \times v} \right|^2} = {\left| u \right|^2}{\left| v \right|^2}$$ and
(b) $$\left( {1 + {{\left| u \right|}^2}} \right)\left( {1 + {{\left| v \right|}^2}} \right) = {\left( {1 - u.v} \right)^2} + {\left| {u + v + \left( {u \times v} \right)} \right|^2}.$$