Key Concepts and Formulas

• Vector Triple Product Identity: $\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C$. This identity is crucial for relating cross products and dot products to individual vectors.
• Properties of Cross Product: $\overrightarrow A \times \overrightarrow B = -(\overrightarrow B \times \overrightarrow A)$ and $\overrightarrow A \times (-\overrightarrow B) = -(\overrightarrow A \times \overrightarrow B)$.
• Dot Product Calculation: For $\overrightarrow u = u_1\widehat i + u_2\widehat j + u_3\widehat k$ and $\overrightarrow v = v_1\widehat i + v_2\widehat j + v_3\widehat k$, $\overrightarrow u \cdot \overrightarrow v = u_1v_1 + u_2v_2 + u_3v_3$.
• Cross Product Calculation: For $\overrightarrow u$ and $\overrightarrow v$, $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$.

Step-by-Step Solution

Step 1: Understand the Given Information and Simplify the Equations We are given: $\overrightarrow a = \widehat j - \widehat k$ $\overrightarrow c = \widehat i - \widehat j - \widehat k$

The conditions are:

1. $\overrightarrow a \times \overrightarrow b + \overrightarrow c = \overrightarrow 0$
2. $\overrightarrow a \cdot \overrightarrow b = 3$

From condition 1, we can isolate the cross product: $\overrightarrow a \times \overrightarrow b = -\overrightarrow c \quad (*)$ Condition 2 directly gives us: $\overrightarrow a \cdot \overrightarrow b = 3 \quad (**)$

Step 2: Apply the Vector Triple Product Identity We want to find $\overrightarrow b$. A common strategy for such problems is to use the vector triple product identity. Let's take the cross product of $\overrightarrow a$ with both sides of equation $(*)$: $\overrightarrow a \times (\overrightarrow a \times \overrightarrow b) = \overrightarrow a \times (-\overrightarrow c)$ Using the vector triple product identity $\overrightarrow A \times (\overrightarrow B \times \overrightarrow C) = (\overrightarrow A \cdot \overrightarrow C)\overrightarrow B - (\overrightarrow A \cdot \overrightarrow B)\overrightarrow C$ with $\overrightarrow A = \overrightarrow a$, $\overrightarrow B = \overrightarrow a$, and $\overrightarrow C = \overrightarrow b$, we get: $(\overrightarrow a \cdot \overrightarrow b)\overrightarrow a - (\overrightarrow a \cdot \overrightarrow a)\overrightarrow b = \overrightarrow a \times (-\overrightarrow c)$ We also know that $\overrightarrow a \times (-\overrightarrow c) = -(\overrightarrow a \times \overrightarrow c)$. So the equation becomes: $(\overrightarrow a \cdot \overrightarrow b)\overrightarrow a - (\overrightarrow a \cdot \overrightarrow a)\overrightarrow b = -(\overrightarrow a \times \overrightarrow c)$

Step 3: Calculate Necessary Vector Components and Products We need to calculate $\overrightarrow a \cdot \overrightarrow b$, $\overrightarrow a \cdot \overrightarrow a$, and $\overrightarrow a \times \overrightarrow c$.

From condition (**), we already have $\overrightarrow a \cdot \overrightarrow b = 3$.

Calculate $\overrightarrow a \cdot \overrightarrow a$: $\overrightarrow a = 0\widehat i + 1\widehat j - 1\widehat k$ $\overrightarrow a \cdot \overrightarrow a = (0)(0) + (1)(1) + (-1)(-1) = 0 + 1 + 1 = 2$ So, $|\overrightarrow a|^2 = 2$.

Calculate $\overrightarrow a \times \overrightarrow c$: $\overrightarrow a \times \overrightarrow c = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 0 & 1 & -1 \\ 1 & -1 & -1 \end{vmatrix}$ $= \widehat i ((1)(-1) - (-1)(-1)) - \widehat j ((0)(-1) - (-1)(1)) + \widehat k ((0)(-1) - (1)(1))$ $= \widehat i (-1 - 1) - \widehat j (0 + 1) + \widehat k (0 - 1)$ $= -2\widehat i - \widehat j - \widehat k$

Step 4: Substitute and Solve for $\overrightarrow b$ Now, substitute the calculated values back into the equation from Step 2: $(3)\overrightarrow a - (2)\overrightarrow b = -(-2\widehat i - \widehat j - \widehat k)$ $3(\widehat j - \widehat k) - 2\overrightarrow b = 2\widehat i + \widehat j + \widehat k$ $3\widehat j - 3\widehat k - 2\overrightarrow b = 2\widehat i + \widehat j + \widehat k$ Rearrange the terms to solve for $2\overrightarrow b$: $2\overrightarrow b = 3\widehat j - 3\widehat k - (2\widehat i + \widehat j + \widehat k)$ $2\overrightarrow b = 3\widehat j - 3\widehat k - 2\widehat i - \widehat j - \widehat k$ Combine like terms: $2\overrightarrow b = -2\widehat i + (3-1)\widehat j + (-3-1)\widehat k$ $2\overrightarrow b = -2\widehat i + 2\widehat j - 4\widehat k$ Divide by 2 to find $\overrightarrow b$: $\overrightarrow b = \frac{1}{2}(-2\widehat i + 2\widehat j - 4\widehat k)$ $\overrightarrow b = -\widehat i + \widehat j - 2\widehat k$

Step 5: Verify the Solution Let's check if our derived $\overrightarrow b = -\widehat i + \widehat j - 2\widehat k$ satisfies the original conditions.

Check $\overrightarrow a \cdot \overrightarrow b = 3$: $\overrightarrow a \cdot \overrightarrow b = (\widehat j - \widehat k) \cdot (-\widehat i + \widehat j - 2\widehat k) = (0)(-1) + (1)(1) + (-1)(-2) = 0 + 1 + 2 = 3$ This condition is satisfied.

Check $\overrightarrow a \times \overrightarrow b = -\overrightarrow c$: First, calculate $\overrightarrow a \times \overrightarrow b$: $\overrightarrow a \times \overrightarrow b = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 0 & 1 & -1 \\ -1 & 1 & -2 \end{vmatrix}$ $= \widehat i ((1)(-2) - (-1)(1)) - \widehat j ((0)(-2) - (-1)(-1)) + \widehat k ((0)(1) - (1)(-1))$ $= \widehat i (-2 + 1) - \widehat j (0 - 1) + \widehat k (0 + 1)$ $= -\widehat i + \widehat j + \widehat k$ Now, calculate $-\overrightarrow c$: $-\overrightarrow c = -(\widehat i - \widehat j - \widehat k) = -\widehat i + \widehat j + \widehat k$ Since $\overrightarrow a \times \overrightarrow b = -\widehat i + \widehat j + \widehat k$ and $-\overrightarrow c = -\widehat i + \widehat j + \widehat k$, this condition is also satisfied.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs during cross product expansion and algebraic manipulation. A single sign mistake can lead to an incorrect final answer.
• Vector Triple Product Application: Ensure you correctly identify $\overrightarrow A, \overrightarrow B, \overrightarrow C$ when applying the identity. Mistakes in this step will propagate through the solution.
• Verification is Key: Always verify your final answer by plugging it back into the original equations. This is a quick way to catch calculation errors.

Summary

This problem is solved by leveraging the vector triple product identity. We used the given conditions to transform the problem into an equation involving the unknown vector $\overrightarrow b$. By calculating the necessary dot and cross products, and performing careful algebraic rearrangement, we successfully derived the vector $\overrightarrow b$. The solution was then verified against the original conditions. The derived vector is $\overrightarrow b = -\widehat i + \widehat j - 2\widehat k$.

The final answer is $\boxed{-\widehat i +\widehat j - 2\widehat k}$.