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Vectors and Its Types

For any two v...

Question

For any two vectors $¯ ¯ ¯ u$ & $¯ ¯ ¯ v$ , prove that
${(¯ ¯ ¯ u . ¯ ¯ ¯ v)}^{2} + {(¯ ¯ ¯ u \times ¯ ¯ ¯ v)}^{2} = {∣ ∣ ¯ ¯ ¯ u ∣ ∣}^{2} {∣ ∣ ¯ ¯ ¯ v ∣ ∣}^{2}$ &
${(1 + ¯ ¯ ¯ u)}^{2} + {(1 + ¯ ¯ ¯ v)}^{2} = {(1 - ¯ ¯ ¯ u . ¯ ¯ ¯ v)}^{2} + {∣ ∣ ¯ ¯ ¯ u + ¯ ¯ ¯ v + (¯ ¯ ¯ u \times ¯ ¯ ¯ v) ∣ ∣}^{2}$

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Solution

Given two vectors, $\to u$ and $\to v$ .
Let the angle between them be $θ$ .
To Prove (i) $(\to u . \to v)^{2} + (\to u \times \to v)^{2} = {| \to u |}^{2} {| \to v |}^{2}$
Dot product can be calculated as
$\to u . \to v = | \to u | | \to v | cos θ$
On squaring the equation we get
$\Rightarrow (\to u . \to v)^{2} = {| \to u |}^{2} {| \to v |}^{2} {cos}^{2} θ$ $\to$ (1)
Cross product can be calculated as
$\to u \times \to v = | \to u | | \to v | sin θ^n$ ,where $^n$ is a unit vector.
On squaring the equation we get
$\Rightarrow$ $(\to u \times \to v)^{2} = {| \to u |}^{2} {| \to v |}^{2} {sin}^{2} θ$ $\to$ (2)
Adding equations 1 and 2 we get,
$(\to u . \to v)^{2} + (\to u \times \to v)^{2} =$ ${| \to u |}^{2} {| \to v |}^{2}$ $({sin}^{2} θ + {cos}^{2} θ)$
$\Rightarrow$ $(\to u . \to v)^{2} + (\to u \times \to v)^{2} = {| \to u |}^{2} {| \to v |}^{2}$
Hence proved.
To Prove (ii) $(\to u + 1)^{2} + (\to v + 1)^{2} = (1 - \to u . \to v)^{2} + {| \to u + \to v + (\to u \times \to v) |}^{2}$
Consider the Right Hand Side of the equation,
$(1 - \to u . \to v)^{2} + {| \to u + \to v + (\to u \times \to v) |}^{2} = 1 + (\to u . \to v)^{2} - 2 (\to u . \to v) + {\to u}^{2} + {\to v}^{2} + (\to u \times \to v)^{2} + 2 (\to u . \to v) + 2 \to u . (\to u \times \to v) + 2 \to v . (\to u \times \to v)$
Note: $\to u . (\to u \times \to v)$ and $\to v . (\to u \times \to v)$ will be equal to zero as $\to u \times \to v$ is perpendicular to both $\to u$ and $\to v$ .
$= 1 + (\to u . \to v)^{2} + {\to u}^{2} + {\to v}^{2} + (\to u \times \to v)^{2}$
On rearranging the terms we get
$= 1 + {\to u}^{2} + {\to v}^{2} + (\to u . \to v)^{2} + (\to u \times \to v)^{2}$
$= 1 + {\to u}^{2} + {\to v}^{2} + {| \to u |}^{2} {| \to v |}^{2}$
$= (1 + {\to u}^{2}) (1 + {\to v}^{2})$
Hence proved.

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