minimum value of vector expression

Question

If $\vec{a}$ and $\vec{b}$ are unit vectors and $\vec{c}$ satisfies $$\vec{c}=-\vec{a}-(\vec{a} \cdot \vec{b})\vec{b} -\vec{a} \times\vec{b}$$ Then the minimum absolute value of $(\vec{a} \times\vec{c} ) \cdot \vec{b}$

my attempt--

$(\vec{a} \times\vec{c} ) \cdot \vec{b}$ can be written down as $\vec{a} \cdot(\vec{c} \times \vec{b})$ and thus i calculated $\vec{c} \times \vec{b}$ and then took the dot product with $\vec{a}$ but could not proceed to find out the minimum value.Please help me with this one.Thanks.

Answer 1

We start to simplify the expression for $(\vec{a} \times\vec{c} ) \cdot \vec{b}$ using properties of cross product:

$$(\vec{a} \times\vec{c} ) \cdot \vec{b}=$$

$$(\vec{a} \times (-\vec{a}-(\vec{a} \cdot \vec{b})\vec{b} -\vec{a} \times\vec{b}) ) \cdot \vec{b}=$$

$$(- \vec{a} \times \vec{a}-(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b}) -\vec{a}\times(\vec{a} \times\vec{b})) \cdot \vec{b}=$$

$$(-(\vec{a} \cdot \vec{b})(\vec{a} \times \vec{b}) - \vec{a}(\vec{a}\cdot \vec{b})+\vec{b}(\vec{a}\cdot \vec{a}) ) \cdot \vec{b}=$$

$$-(\vec{a} \cdot \vec{b})((\vec{a} \times \vec{b})\cdot \vec{b}) - (\vec{a}\cdot \vec{b})(\vec{a}\cdot \vec{b})+(\vec{b}\cdot \vec{b}) (\vec{a}\cdot \vec{a})=$$

$$(\vec{b}\cdot \vec{b}) (\vec{a}\cdot \vec{a}) - (\vec{a}\cdot \vec{b})(\vec{a}\cdot \vec{b}).$$

The last expression is the left hand side of Cauchy-Shwarz inequality for vectors $\vec{a}$ and $\vec{b}$, so it is at least zero and it is zero iff the vectors $\vec{a}$ and $\vec{b}$ are parallel, in case of their unit length iff $\vec{a}=\pm \vec{b}$.

Answer 2

Hint: $$(\vec{a} \times\vec{c} ) \cdot \vec{b} = -(\vec{c} \times\vec{a} ) \cdot \vec{b} = \vec{a}\cdot(\vec{a}\times\vec{b})+(\vec{a}\cdot\vec{b})\vec{b}\times(\vec{a}\times\vec{b}) + (\vec{a}\times\vec{b})\cdot(\vec{a}\times\vec{b}).$$ Simplify the first two terms.

Answer 3

Hint: you need to use the identity \begin{equation} A\times(B\times C)=(A \cdot C) B- (A \cdot B) C\, . \end{equation}