Key Concepts and Formulas

1. Cross Product Property for Parallel Vectors: If $\vec{u} \times \vec{v} = \vec{0}$ and $\vec{u} \neq \vec{0}$, then $\vec{v}$ is parallel to $\vec{u}$, meaning $\vec{v} = k\vec{u}$ for some scalar $k$.
2. Distributive Property of Cross Product: $\vec{a} \times (\vec{b} - \vec{c}) = \vec{a} \times \vec{b} - \vec{a} \times \vec{c}$.
3. Distributive Property of Dot Product: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$.
4. Dot Product of a Vector with Itself: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$.
5. Dot Product Commutativity: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.
6. Component-wise Dot Product: If $\vec{u} = u_1\widehat{i} + u_2\widehat{j} + u_3\widehat{k}$ and $\vec{v} = v_1\widehat{i} + v_2\widehat{j} + v_3\widehat{k}$, then $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3$.

Step-by-Step Solution

1. Analyze the Given Information and Goal

We are given two vectors, $\overrightarrow a = \widehat i - 2\widehat j + \widehat k$ and $\overrightarrow b = \widehat i - \widehat j + \widehat k$. We are also given two conditions involving an unknown vector $\overrightarrow c$: (i) $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a$ (ii) $\overrightarrow c \cdot \overrightarrow a = 0$ Our goal is to find the value of $\overrightarrow c \cdot \overrightarrow b$.

2. Simplify the Cross Product Condition

The first condition, $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a$, can be rearranged.

• Rearrange the equation: Subtract $\overrightarrow b \times \overrightarrow a$ from both sides to get: $\overrightarrow b \times \overrightarrow c - \overrightarrow b \times \overrightarrow a = \overrightarrow 0$ Why: This step is taken to group terms involving the cross product, enabling the use of its distributive property.

• Apply the distributive property of the cross product: Factor out $\overrightarrow b$ from the left side: $\overrightarrow b \times (\overrightarrow c - \overrightarrow a) = \overrightarrow 0$ Why: This transformation is crucial because it puts the equation in the form $\vec{u} \times \vec{v} = \vec{0}$.

• Deduce the parallelism: Since $\overrightarrow b \times (\overrightarrow c - \overrightarrow a) = \overrightarrow 0$, and we know $\overrightarrow b$ is not the zero vector (as its components are 1, -1, 1), the vector $(\overrightarrow c - \overrightarrow a)$ must be parallel to $\overrightarrow b$. Why: The cross product of two non-zero vectors is zero if and only if they are parallel.

• Express $(\overrightarrow c - \overrightarrow a)$ in terms of $\overrightarrow b$: If $(\overrightarrow c - \overrightarrow a)$ is parallel to $\overrightarrow b$, then $(\overrightarrow c - \overrightarrow a)$ can be written as a scalar multiple of $\overrightarrow b$. Let this scalar be $\lambda$. $\overrightarrow c - \overrightarrow a = \lambda \overrightarrow b$ Why: Introducing a scalar $\lambda$ allows us to represent the relationship between the parallel vectors and will help us solve for the unknown vector $\overrightarrow c$.

• Express $\overrightarrow c$ in terms of $\overrightarrow a$ and $\overrightarrow b$: Rearranging the equation, we get: $\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b$ Why: This equation provides a general form for $\overrightarrow c$ that satisfies the first given condition. We will use the second condition to determine the specific value of $\lambda$.

3. Use the Dot Product Condition to Find the Scalar $\lambda$

The second condition is $\overrightarrow c \cdot \overrightarrow a = 0$. We substitute the expression for $\overrightarrow c$ from Step 2 into this condition.

• Substitute the expression for $\overrightarrow c$: $(\overrightarrow a + \lambda \overrightarrow b) \cdot \overrightarrow a = 0$ Why: By substituting the general form of $\overrightarrow c$, we obtain an equation that involves only known vectors and the unknown scalar $\lambda$.

• Apply the distributive property of the dot product: $\overrightarrow a \cdot \overrightarrow a + \lambda (\overrightarrow b \cdot \overrightarrow a) = 0$ Why: The dot product distributes over vector addition, allowing us to expand the expression.

• Calculate the necessary dot products and magnitudes: We need $\overrightarrow a \cdot \overrightarrow a$ and $\overrightarrow b \cdot \overrightarrow a$. $\overrightarrow a \cdot \overrightarrow a = |\overrightarrow a|^2 = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6$ $\overrightarrow b \cdot \overrightarrow a = \overrightarrow a \cdot \overrightarrow b = (1)(1) + (-2)(-1) + (1)(1) = 1 + 2 + 1 = 4$ Why: Calculating these values allows us to substitute them into the equation and solve for $\lambda$.

• Solve for $\lambda$: Substitute the calculated values into the equation from the distributive property: $6 + \lambda (4) = 0$ $4\lambda = -6$ $\lambda = -\frac{6}{4} = -\frac{3}{2}$ Why: We have now found the specific scalar value $\lambda$ that makes $\overrightarrow c$ satisfy both given conditions.

4. Calculate the Required Dot Product $\overrightarrow c \cdot \overrightarrow b$

We need to find $\overrightarrow c \cdot \overrightarrow b$. We know that $\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b$, and we have found $\lambda = -\frac{3}{2}$.

• Substitute the expression for $\overrightarrow c$ and the value of $\lambda$: $\overrightarrow c \cdot \overrightarrow b = (\overrightarrow a + \lambda \overrightarrow b) \cdot \overrightarrow b$ $\overrightarrow c \cdot \overrightarrow b = (\overrightarrow a - \frac{3}{2} \overrightarrow b) \cdot \overrightarrow b$ Why: We substitute the expression for $\overrightarrow c$ into the dot product we need to compute.

• Apply the distributive property of the dot product: $\overrightarrow c \cdot \overrightarrow b = \overrightarrow a \cdot \overrightarrow b - \frac{3}{2} (\overrightarrow b \cdot \overrightarrow b)$ Why: The dot product distributes over vector subtraction, allowing us to separate the terms.

• Calculate the necessary dot products and magnitudes: We already calculated $\overrightarrow a \cdot \overrightarrow b = 4$ in Step 3. Now we need $\overrightarrow b \cdot \overrightarrow b$. $\overrightarrow b \cdot \overrightarrow b = |\overrightarrow b|^2 = (1)^2 + (-1)^2 + (1)^2 = 1 + 1 + 1 = 3$ Why: These values are needed to complete the calculation of $\overrightarrow c \cdot \overrightarrow b$.

• Compute the final result: Substitute the calculated values into the expanded dot product expression: $\overrightarrow c \cdot \overrightarrow b = 4 - \frac{3}{2} (3)$ $\overrightarrow c \cdot \overrightarrow b = 4 - \frac{9}{2}$ $\overrightarrow c \cdot \overrightarrow b = \frac{8}{2} - \frac{9}{2}$ $\overrightarrow c \cdot \overrightarrow b = -\frac{1}{2}$ Why: This is the final value of the required dot product, obtained by combining the results of our vector manipulations.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with signs and fractions when performing vector addition, subtraction, and scalar multiplication.
• Misinterpreting Cross Product: Remember that $\vec{u} \times \vec{v} = \vec{0}$ implies parallelism, not necessarily equality or being the zero vector.
• Efficiency: Instead of finding the explicit components of $\overrightarrow c$ first, it is more efficient to use its general form $\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b$ directly in the final dot product calculation.

Summary

The problem is solved by first using the cross product condition $\overrightarrow b \times \overrightarrow c = \overrightarrow b \times \overrightarrow a$ to deduce that $\overrightarrow c - \overrightarrow a$ is parallel to $\overrightarrow b$, leading to the expression $\overrightarrow c = \overrightarrow a + \lambda \overrightarrow b$. Subsequently, the dot product condition $\overrightarrow c \cdot \overrightarrow a = 0$ is used to form an equation involving the scalar $\lambda$, which is solved to be $\lambda = -3/2$. Finally, the required dot product $\overrightarrow c \cdot \overrightarrow b$ is computed by substituting the expression for $\overrightarrow c$ and the value of $\lambda$.

The final answer is $\boxed{{ - {1 \over 2}}}$, which corresponds to option (A).