Dot and Cross Products

General Coordinate Free Result:

The vector $\vec{B}$ can be expressed in terms of $\vec{A}$ , $\vec{A} \cdot\vec{B}$ , and $\vec{A} \times \vec{B}$ as follows: $\vec{B} = \frac{ ( \vec{A} \cdot \vec{B} ) \vec{A} \ - \ \vec{A} \times (\vec{A} \times \vec{B})}{\vec{A} \cdot \vec{A}}$

---

Derivation #1:

Note that $\vec{A}$ , $\vec{A} \times \vec{B}$ , and $\vec{A} \times (\vec{A} \times \vec{B})$ are three mutually orthogonal vectors. Thus, any vector can be written as a linear combination of these three basis vectors (as long as $\vec{A} \times \vec{B} \neq 0$ ). Let us do so with the specific vector $\vec{B}$ . $\vec{B} = \color{#D61F06}p \color{#333333}\vec{A} + \color{#D61F06}q \color{#333333}\vec{A}\times \vec{B} + \color{#D61F06}r \color{#333333} \vec{A} \times (\vec{A} \times \vec{B}) \qquad (*)$ We now try to express the coefficients $\color{#D61F06}p$ , $\color{#D61F06}q$ , and $\color{#D61F06}r$ in terms of the known information about $\vec{A}$ and $\vec{A} \cdot \vec{B}$ .

1. Take the dot product of $\vec{A}$ with $(*)$ : $\vec{A} \cdot \vec{B} = \color{#D61F06}p \color{#333333} \vec{A} \cdot \vec{A}$ , so $\boxed { \color{#D61F06}p \color{#333333} = \frac{ \vec{A} \cdot \vec{B}}{\vec{A} \cdot \vec{A}} }$ .

2. Take the dot product of $\vec{A} \times \vec{B}$ with $(*)$ : $( \vec{A} \times \vec{B} ) \cdot \vec{B} = \color{#D61F06}q \color{#333333} ( \vec{A} \times \vec{B} ) \cdot ( \vec{A} \times \vec{B} )$ . But the left hand side is zero since $\vec{A} \times \vec{B}$ is perpendicular to $\vec{B}$ . So $\boxed{ \color{#D61F06}q \color{#333333} = 0 }$ .

3. Take the cross product of $\vec{A}$ with $(*)$ . With the middle $\color{#D61F06}q$ term dropped, there is only one term remaining on the right hand side: $\vec{A} \times \vec{B} = \color{#D61F06}r \color{#333333} \vec{A} \times ( \vec{A} \times (\vec{A} \times \vec{B}) )$ . The right hand side looks ugly, but isn't too bad. The outer two cross products are of perpendicular vectors, so they just scale the vector $\vec{A} \times \vec{B}$ by the magnitude of $\vec{A}$ twice: $|\vec{A} \times ( \vec{A} \times (\vec{A} \times \vec{B}))| = |\vec{A}|^2 |\vec{A} \times \vec{B}|$ . And consideration of right hand rule makes it clear that the direction is antiparallel to the original vector $\vec{A} \times \vec{B}$ . Therefore, this third equation becomes $\vec{A} \times \vec{B} = -\color{#D61F06}r \color{#333333} |\vec{A}|^2 (\vec{A} \times \vec{B})$ . (These quick claims can be made more explicit by considering the vector triple product identity. See Derivation #2.) Since $\vec{A} \times \vec{B} \neq 0$ , we conclude $\boxed{ \color{#D61F06}r \color{#333333} = - \frac{1}{| \vec{A} |^2} = - \frac{1}{\vec{A} \cdot \vec{A}} }$ .

This proves the general result stated above.

---

Derivation #2:

A well known identity for the vector triple product is $\vec{A} \times ( \vec{B} \times \vec{C} ) = \vec{B} ( \vec{A} \cdot \vec{C} ) - \vec{C} ( \vec{A} \cdot \vec{B} )$ . Substitute $\vec{A}$ for $\vec{B}$ and $\vec{B}$ for $\vec{C}$ to obtain $\vec{A} \times ( \vec{A} \times \vec{B} ) = \vec{A} ( \vec{A} \cdot \vec{B} ) - \vec{B} ( \vec{A} \cdot \vec{A} )$ Solve this equation for the vector $\vec{B}$ on the right hand side. $\vec{B} ( \vec{A} \cdot \vec{A} ) = \vec{A} ( \vec{A} \cdot \vec{B} ) - \vec{A} \times ( \vec{A} \times \vec{B} )$ $\vec{B} = \frac{\vec{A} ( \vec{A} \cdot \vec{B} ) - \vec{A} \times ( \vec{A} \times \vec{B} )}{\vec{A} \cdot \vec{A}} \qquad \text{if } \vec{A} \neq 0$

---

Application:

To apply this general result to the specific problem given here, we have a couple terms to evaluate.

1. $( \vec{A} \cdot \vec{B} ) \vec{A} = (11)( 2\hat{i} + \hat{j} + 5\hat{k}) = 22\hat{i} + 11\hat{j} + 55\hat{k}$

2. $\vec{A} \times ( \vec{A} \times \vec{B} ) = ( 2\hat{i} + \hat{j} + 5\hat{k} ) \times ( -13\hat{i} - 9\hat{j} + 7\hat{k} ) = 52\hat{i} - 79\hat{j} - 5\hat{k}$

3. $\vec{A} \cdot \vec{A} = ( 2\hat{i} + \hat{j} + 5\hat{k} ) \cdot ( 2\hat{i} + \hat{j} + 5\hat{k} ) = 2^2 +1^2 +5^2 = 30$

Plugging it all in gives us the Cartesian components of $\vec{B}$ . $\begin{aligned} \vec{B} & = \frac{\vec{A} ( \vec{A} \cdot \vec{B} ) - \vec{A} \times ( \vec{A} \times \vec{B} )}{\vec{A} \cdot \vec{A}} \\ & = \frac{ (22\hat{i} + 11\hat{j} + 55\hat{k}) - (52\hat{i} - 79\hat{j} - 5\hat{k}) }{30} \\ & = \frac{ -30\hat{i} + 90\hat{j} + 60\hat{k} }{30} \\ & = -\hat{i} + 3\hat{j} + 2\hat{k} \end{aligned}$ So the requested product is $(-1)(3)(2) = \color{#D61F06}\boxed{-6}$ .

We have 4 equations and 3 variables. Dot product: $\vec{A} \cdot \vec{B}=2x+y+5z=11$ and the Cross product: $z-5y=-13, 5x-2z=-9, 2y-x=7$ The cross product equations can be put in matrix form. $\left[ \begin{array}{ccc|c} 0 & -5 & 1 & -13 \\ 5 & 0 & -2 & -9 \\ -1 & 2 & 0 & 7 \\ \end{array} \right] \sim \left[ \begin{array}{ccc|c} 0 & 5 & -1 & 13 \\ 0 & 10 & -2 & 26 \\ 1 & -2 & 0 & -7 \\ \end{array} \right]$ The first two equations are dependent, we replace one of them with the equation from the dot product and continue to to solve:

$\left[ \begin{array}{ccc|c} 1 & -2 & 0 & -7 \\ 0 & 5 & -1 & 13 \\ 2 & 1 & 5 &11 \\ \end{array} \right] \sim \left[ \begin{array}{ccc|c} 1 & -2 & 0 & -7\\ 0 & 5 & -1 & 13 \\ 0 & 5 & 5 & 25 \\ \end{array} \right] \sim \left[ \begin{array}{ccc|c} 1 & -2 & 0 & -7\\ 0 & 0 & -6 & -12 \\ 0 & 1 & 1 & 5 \\ \end{array} \right] \sim \left[ \begin{array}{ccc|c} 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & 2 \\ 0 & 1 & 0 & 3 \\ \end{array} \right]$ So $\vec{B}=(-1, 3,2)$ and the answer is $\boxed{-6}$