1. ## Proving this equation?

l a x b l^2 + (a dot b)^2 = lal^2 lbl^2

a and b are vectors

well i know that a cross b = lallblsin theta and a dot b = lallblcos theta but i don't know how to proceed from here...

how would i go about proving this using the definitions of a cross b and a dot b? why is this equation equivalent to the Pythagorean identity?

2. $\displaystyle | \vec A \times \vec B |^2 + \left ( \vec A \cdot \vec B \right ) ^2 = \left ( |\vec A||\vec B|\sin{\theta} \right ) ^2 + \left ( |\vec A||\vec B|\cos{\theta} \right ) ^2$

Factor $\displaystyle |\vec A|^2|\vec B|^2$.

$\displaystyle |\vec A|^2|\vec B|^2\sin^2{\theta} + |\vec A|^2|\vec B|^2\cos^2{\theta} = |\vec A|^2|\vec B|^2 \left(\sin^2{\theta} + \cos^2{\theta} \right)$

Use the trigonometric identity $\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1$.

$\displaystyle |\vec A|^2|\vec B|^2 \left(\sin^2{\theta} + \cos^2{\theta} \right) = |\vec A|^2|\vec B|^2$

3. You're almost there...

$\displaystyle |\mathbf{a}\times\mathbf{b}| + (\mathbf{a}\cdot\mathbf{b})^2 = (|\mathbf{a}||\mathbf{b}|\sin{\theta})^2 + (|\mathbf{a}||\mathbf{b}|\cos{\theta})^2$

$\displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2\sin^2{\theta} + |\mathbf{a}|^2|\mathbf{b}|^2\cos^2{\theta}$

$\displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2(\sin^2{\theta} + \cos^2{\theta})$

$\displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2(1)$

$\displaystyle = |\mathbf{a}|^2|\mathbf{b}|^2$.

4. $\mid a X b\mid=\mid a\mid\mid b\mid sin\theta$...........1

$a.b=\mid a\mid \mid b \mid cos\theta$................2

add the square of both the equation and use $cos^2\theta+sin^2\theta=1$