Key Concepts and Formulas

• Right-Handed System of Vectors: Three vectors $\overrightarrow u, \overrightarrow v, \overrightarrow w$ form a right-handed system if, when you curl the fingers of your right hand from $\overrightarrow u$ to $\overrightarrow v$ through the smaller angle, your thumb points in the direction of $\overrightarrow w$. Mathematically, this implies $\overrightarrow u \times \overrightarrow v$ is in the same direction as $\overrightarrow w$. For a cyclic order $(\overrightarrow a, \overrightarrow c, \overrightarrow b)$ to form a right-handed system, the relationship $\overrightarrow a \times \overrightarrow c = k\overrightarrow b$ for some $k>0$ must hold. A more direct implication for finding $\overrightarrow c$ is that $\overrightarrow b \times \overrightarrow a$ will be in the direction of $\overrightarrow c$.
• Vector Cross Product: The cross product of two vectors $\overrightarrow P = P_x\widehat i + P_y\widehat j + P_z\widehat k$ and $\overrightarrow Q = Q_x\widehat i + Q_y\widehat j + Q_z\widehat k$ is given by the determinant: $\overrightarrow P \times \overrightarrow Q = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ P_x & P_y & P_z \\ Q_x & Q_y & Q_z \end{vmatrix}$ Expanding this determinant gives: $\overrightarrow P \times \overrightarrow Q = (P_y Q_z - P_z Q_y)\widehat i - (P_x Q_z - P_z Q_x)\widehat j + (P_x Q_y - P_y Q_x)\widehat k$

Step-by-Step Solution

Step 1: Understand the Right-Handed System Property We are given that the vectors $\overrightarrow a, \overrightarrow c, \overrightarrow b$ form a right-handed system. This means that if we consider the vectors in this specific order, their cross product relationship follows a cyclic pattern. Specifically, for the order $(\overrightarrow a, \overrightarrow c, \overrightarrow b)$ to be right-handed, the cross product of the first two vectors, $\overrightarrow a \times \overrightarrow c$, must be in the same direction as the third vector, $\overrightarrow b$. That is, $\overrightarrow a \times \overrightarrow c = k\overrightarrow b$ for some positive scalar $k$. However, to find $\overrightarrow c$ directly, we can use the property that if $(\overrightarrow a, \overrightarrow c, \overrightarrow b)$ form a right-handed system, then the cross product of the last two vectors in the cyclic order, $\overrightarrow b \times \overrightarrow a$, will be in the same direction as the first vector, $\overrightarrow a$, and similarly, $\overrightarrow c \times \overrightarrow b$ will be in the direction of $\overrightarrow a$. The most useful relationship for finding $\overrightarrow c$ is derived from the cyclic permutation: $\overrightarrow a \times \overrightarrow c = k\overrightarrow b$. If we consider the vectors in the order $(\overrightarrow b, \overrightarrow a, \overrightarrow c)$ then $\overrightarrow b \times \overrightarrow a$ would be in the direction of $\overrightarrow c$. Thus, we can express $\overrightarrow c$ as: $\overrightarrow c = \overrightarrow b \times \overrightarrow a$ This is because if you place $\overrightarrow b$ along the positive y-axis (as $\widehat j$) and $\overrightarrow a$ can be any vector, then rotating from $\overrightarrow b$ to $\overrightarrow a$ (in the correct sense for a right-handed system) should result in $\overrightarrow c$. The cross product $\overrightarrow b \times \overrightarrow a$ captures this directional relationship.

Step 2: Define the Given Vectors We are given:

• $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$, which can be written in component form as $(x, y, z)$.
• $\overrightarrow b = \widehat j$, which can be written in component form as $(0, 1, 0)$.

Step 3: Calculate the Cross Product $\overrightarrow b \times \overrightarrow a$ We will use the determinant formula for the cross product: $\overrightarrow c = \overrightarrow b \times \overrightarrow a = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ b_x & b_y & b_z \\ a_x & a_y & a_z \end{vmatrix}$ Substitute the components of $\overrightarrow b = (0, 1, 0)$ and $\overrightarrow a = (x, y, z)$: $\overrightarrow c = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 0 & 1 & 0 \\ x & y & z \end{vmatrix}$

Step 4: Evaluate the Determinant Expand the determinant along the first row: $\overrightarrow c = \widehat i \begin{vmatrix} 1 & 0 \\ y & z \end{vmatrix} - \widehat j \begin{vmatrix} 0 & 0 \\ x & z \end{vmatrix} + \widehat k \begin{vmatrix} 0 & 1 \\ x & y \end{vmatrix}$ Calculate each $2 \times 2$ determinant:

• The $\widehat i$ component: $(1 \cdot z - 0 \cdot y) = z - 0 = z$
• The $\widehat j$ component: $(0 \cdot z - 0 \cdot x) = 0 - 0 = 0$. Remember the negative sign in the formula: $- (0) = 0$.
• The $\widehat k$ component: $(0 \cdot y - 1 \cdot x) = 0 - x = -x$

Step 5: Combine the Components to Find $\overrightarrow c$ Summing the components, we get: $\overrightarrow c = z\widehat i - 0\widehat j + (-x)\widehat k$ $\overrightarrow c = z\widehat i - x\widehat k$

Common Mistakes & Tips

• Order of Cross Product: The definition of a right-handed system is sensitive to the order of the vectors. Ensure you correctly identify which cross product yields $\overrightarrow c$. For $(\overrightarrow a, \overrightarrow c, \overrightarrow b)$ to be right-handed, it implies $\overrightarrow a \times \overrightarrow c$ is in the direction of $\overrightarrow b$. This does not directly give $\overrightarrow c$. However, the cyclic nature implies that $\overrightarrow b \times \overrightarrow a$ is in the direction of $\overrightarrow c$.
• Determinant Calculation Signs: Be very careful with the signs when expanding the determinant. The $\widehat j$ term always has a negative sign in the expansion formula.
• Vector Components: Double-check that you are placing the components of the correct vectors in the correct rows of the determinant. The first vector in the cross product goes in the second row, and the second vector goes in the third row.

Summary

To solve this problem, we leveraged the definition of a right-handed system of vectors. If $\overrightarrow a, \overrightarrow c, \overrightarrow b$ form a right-handed system in that order, then the cross product $\overrightarrow b \times \overrightarrow a$ yields a vector in the direction of $\overrightarrow c$. We then proceeded to calculate this cross product using the standard determinant method, substituting the given components of $\overrightarrow a$ and $\overrightarrow b$. The evaluation of the determinant provided the vector $\overrightarrow c$.

The final answer is $\boxed{z\widehat i - x\widehat k}$.