# Question #8f5e6

Jul 31, 2016

$\frac{\vec{d} \times \left(\vec{a} \times \vec{d}\right)}{| \vec{d} {|}^{2}}$

bac-cab rule for triple vec product
$= \frac{\vec{a} \left(\vec{d} \cdot \vec{d}\right) - \vec{d} \left(\vec{d} \cdot \vec{a}\right)}{| \vec{d} {|}^{2}}$

As $\vec{d} \cdot \vec{d} = | \vec{d} {|}^{2}$
$= \vec{a} - \frac{\vec{d} \left(\vec{d} \cdot \vec{b} + \vec{d} \cdot \vec{c}\right)}{| \vec{d} {|}^{2}}$

As $\vec{c} \cdot \vec{d} = \vec{0} = \vec{d} \cdot \vec{c}$

And as $\vec{b} \times \vec{d} = \vec{0} = | \vec{b} | | \vec{c} | \sin \theta \setminus \hat{n} \implies \theta = k \pi , k \in m a t h c a l \left(Z\right)$ so $\vec{b} \cdot \vec{d} = \pm | \vec{b} | | \vec{d} |$

$= \vec{a} \pm \frac{\vec{d} \left(\left\mid \vec{d} \right\mid \left\mid \vec{b} \right\mid\right)}{| \vec{d} {|}^{2}}$

$= \vec{a} \pm \frac{\vec{d} \left\mid \vec{b} \right\mid}{| \vec{d} |}$

$= \vec{a} \pm \hat{d} \left\mid \vec{b} \right\mid$

just doesn't feel right but that's where i get it to come out