Key Concepts and Formulas

1. Vector Operations: The problem relies on the dot product ($\cdot$) and cross product ($\times$) of vectors.
   • Dot Product: $\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta$, or component-wise: $(a_x \widehat{i} + a_y \widehat{j} + a_z \widehat{k}) \cdot (b_x \widehat{i} + b_y \widehat{j} + b_z \widehat{k}) = a_x b_x + a_y b_y + a_z b_z$. The dot product is distributive: $\overrightarrow{u} \cdot (\overrightarrow{v} + \overrightarrow{w}) = \overrightarrow{u} \cdot \overrightarrow{v} + \overrightarrow{u} \cdot \overrightarrow{w}$.
   • Cross Product: $\overrightarrow{a} \times \overrightarrow{b}$ results in a vector perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$. A crucial property is that $(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{b} = 0$ and $(\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{a} = 0$.
2. Standard Basis Vectors: $\widehat{i}$, $\widehat{j}$, $\widehat{k}$ are mutually orthogonal unit vectors. Their dot products are: $\widehat{i} \cdot \widehat{i} = \widehat{j} \cdot \widehat{j} = \widehat{k} \cdot \widehat{k} = 1$, and $\widehat{i} \cdot \widehat{j} = \widehat{j} \cdot \widehat{k} = \widehat{k} \cdot \widehat{i} = 0$.

Step-by-Step Solution

Step 1: Simplify the Target Expression Our goal is to evaluate the expression $E = (\overrightarrow b - \overrightarrow a )\,.\,(\widehat k - \widehat j) + (\overrightarrow b + \overrightarrow a )\,.\,(\widehat i - \widehat k)$. We will use the distributive property of the dot product. $E = (\overrightarrow b \cdot \widehat k - \overrightarrow b \cdot \widehat j - \overrightarrow a \cdot \widehat k + \overrightarrow a \cdot \widehat j) + (\overrightarrow b \cdot \widehat i - \overrightarrow b \cdot \widehat k + \overrightarrow a \cdot \widehat i - \overrightarrow a \cdot \widehat k)$ Group terms involving $\overrightarrow b$ and $\overrightarrow a$: $E = (\overrightarrow b \cdot \widehat k - \overrightarrow b \cdot \widehat k) + (\overrightarrow b \cdot \widehat i - \overrightarrow b \cdot \widehat j) + (\overrightarrow a \cdot \widehat i + \overrightarrow a \cdot \widehat j - 2\overrightarrow a \cdot \widehat k)$ The terms $\overrightarrow b \cdot \widehat k$ cancel out. $E = \overrightarrow b \cdot (\widehat i - \widehat j) + \overrightarrow a \cdot (\widehat i + \widehat j - 2\widehat k)$ We are given $\overrightarrow b = \widehat i + \widehat j + \lambda \widehat k$. Let's compute the first part of the expression: $\overrightarrow b \cdot (\widehat i - \widehat j) = (\widehat i + \widehat j + \lambda \widehat k) \cdot (\widehat i - \widehat j)$ Using the properties of dot products with standard basis vectors: $= (1)(1) + (1)(-1) + (\lambda)(0) = 1 - 1 + 0 = 0$ This simplifies the target expression significantly: $E = 0 + \overrightarrow a \cdot (\widehat i + \widehat j - 2\widehat k) = \overrightarrow a \cdot (\widehat i + \widehat j - 2\widehat k)$ Let $\overrightarrow a = a_x \widehat i + a_y \widehat j + a_z \widehat k$. Then, $E = a_x(1) + a_y(1) + a_z(-2) = a_x + a_y - 2a_z$. Our task is now to find the components of $\overrightarrow a$.

Step 2: Determine the Value of $\lambda$ We are given $\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat k$. Let $\overrightarrow d = \overrightarrow a \times \overrightarrow b$. A key property of the cross product is that it is orthogonal to both vectors involved. Therefore, $(\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow b = 0$. We are given $\overrightarrow b = \widehat i + \widehat j + \lambda \widehat k$ and $\overrightarrow d = 13\widehat i - \widehat j - 4\widehat k$. $(13\widehat i - \widehat j - 4\widehat k) \cdot (\widehat i + \widehat j + \lambda \widehat k) = 0$ $(13)(1) + (-1)(1) + (-4)(\lambda) = 0$ $13 - 1 - 4\lambda = 0$ $12 - 4\lambda = 0$ $4\lambda = 12$ $\lambda = 3$ So, $\overrightarrow b = \widehat i + \widehat j + 3\widehat k$.

Step 3: Determine the Components of $\overrightarrow a$ Let $\overrightarrow a = a_x \widehat i + a_y \widehat j + a_z \widehat k$. We use the two given conditions to set up equations for $a_x, a_y, a_z$.

Condition 1: $\overrightarrow a \times \overrightarrow b = 13\widehat i - \widehat j - 4\widehat k$ Substitute $\overrightarrow a$ and $\overrightarrow b = \widehat i + \widehat j + 3\widehat k$ into the cross product formula: $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ a_x & a_y & a_z \\ 1 & 1 & 3 \end{vmatrix} = (3a_y - a_z)\widehat{i} - (3a_x - a_z)\widehat{j} + (a_x - a_y)\widehat{k}$ Equating this to $13\widehat i - \widehat j - 4\widehat k$:

1. $3a_y - a_z = 13$
2. $-(3a_x - a_z) = -1 \implies 3a_x - a_z = 1$
3. $a_x - a_y = -4$

Condition 2: $\overrightarrow a \cdot \overrightarrow b = -21$ Substitute $\overrightarrow a$ and $\overrightarrow b = \widehat i + \widehat j + 3\widehat k$ into the dot product formula: $(a_x \widehat i + a_y \widehat j + a_z \widehat k) \cdot (\widehat i + \widehat j + 3\widehat k) = -21$ 4. $a_x + a_y + 3a_z = -21$

Now we solve the system of equations (1), (2), (3), and (4). From equation (3), $a_y = a_x + 4$. From equation (2), $a_z = 3a_x - 1$.

Substitute these into equation (1) to check for consistency: $3(a_x + 4) - (3a_x - 1) = 13$ $3a_x + 12 - 3a_x + 1 = 13$ $13 = 13$. The first three equations are consistent.

Now substitute $a_y$ and $a_z$ into equation (4): $a_x + (a_x + 4) + 3(3a_x - 1) = -21$ $2a_x + 4 + 9a_x - 3 = -21$ $11a_x + 1 = -21$ $11a_x = -22$ $a_x = -2$

Now find $a_y$ and $a_z$: $a_y = a_x + 4 = -2 + 4 = 2$ $a_z = 3a_x - 1 = 3(-2) - 1 = -6 - 1 = -7$ Thus, $\overrightarrow a = -2\widehat i + 2\widehat j - 7\widehat k$.

Step 4: Evaluate the Simplified Target Expression We found that the target expression is $E = a_x + a_y - 2a_z$. Substitute the values $a_x = -2$, $a_y = 2$, and $a_z = -7$: $E = (-2) + (2) - 2(-7)$ $E = 0 + 14$ $E = 14$

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with signs and arithmetic when expanding dot and cross products and solving systems of equations.
• Forgetting Properties: The property $(\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow b = 0$ is a shortcut for finding unknown scalars in $\overrightarrow b$. Misapplying or forgetting this property can lead to a much more complex solution.
• Simplification: Always look for opportunities to simplify the expression before diving into component calculations. The cancellation of $\overrightarrow b \cdot (\widehat i - \widehat j)$ was key here.

Summary The problem involves simplifying a vector expression and finding the components of an unknown vector $\overrightarrow a$. We first simplified the target expression to $\overrightarrow a \cdot (\widehat i + \widehat j - 2\widehat k)$. Then, we used the orthogonality property of the cross product, $(\overrightarrow a \times \overrightarrow b) \cdot \overrightarrow b = 0$, to find the value of $\lambda$ in $\overrightarrow b$. With $\overrightarrow b$ determined, we used the given cross product and dot product conditions to form a system of linear equations for the components of $\overrightarrow a$. Solving this system yielded $\overrightarrow a$. Finally, we substituted the components of $\overrightarrow a$ into the simplified target expression to obtain the numerical value.

The final answer is $\boxed{14}$.