Key Concepts and Formulas

1. Cross Product Property for Parallel Vectors: If $\overrightarrow{u} \times \overrightarrow{v} = \overrightarrow{0}$ and $\overrightarrow{v} \neq \overrightarrow{0}$, then $\overrightarrow{u}$ is parallel to $\overrightarrow{v}$, which means $\overrightarrow{u} = \lambda \overrightarrow{v}$ for some scalar $\lambda$.
2. Properties of Vector Operations:
   • Anticommutativity of Cross Product: $\overrightarrow{a} \times \overrightarrow{b} = -(\overrightarrow{b} \times \overrightarrow{a})$
   • Distributivity of Cross Product: $\overrightarrow{a} \times (\overrightarrow{b} + \overrightarrow{c}) = \overrightarrow{a} \times \overrightarrow{b} + \overrightarrow{a} \times \overrightarrow{c}$
   • Distributivity of Dot Product: $\overrightarrow{a} \cdot (\overrightarrow{b} + \overrightarrow{c}) = \overrightarrow{a} \cdot \overrightarrow{b} + \overrightarrow{a} \cdot \overrightarrow{c}$
   • Scalar Multiplication in Dot Product: $(k\overrightarrow{a}) \cdot \overrightarrow{b} = k(\overrightarrow{a} \cdot \overrightarrow{b})$

Step-by-Step Solution

We are given the vectors: $\overrightarrow a = \widehat i + 2\widehat k$ $\overrightarrow b = \widehat i + \widehat j + \widehat k$ $\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k$

And the conditions for vector $\overrightarrow r$:

1. $\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0$
2. $\overrightarrow r \,.\,\overrightarrow a = 0$

We need to find $\overrightarrow r \,.\,\overrightarrow c$.

Step 1: Simplify the cross product condition to find the general form of $\overrightarrow r$.

The first condition is $\overrightarrow r \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = \overrightarrow 0$. Using the anticommutativity property of the cross product, we can rewrite $\overrightarrow b \times \overrightarrow c$ as $-(\overrightarrow c \times \overrightarrow b)$. $\overrightarrow r \times \overrightarrow b - (\overrightarrow c \times \overrightarrow b) = \overrightarrow 0$ Now, applying the distributive property of the cross product, we can factor out $\overrightarrow b$: $(\overrightarrow r - \overrightarrow c) \times \overrightarrow b = \overrightarrow 0$ This equation implies that the vector $(\overrightarrow r - \overrightarrow c)$ is parallel to the vector $\overrightarrow b$, since their cross product is the zero vector and $\overrightarrow b \neq \overrightarrow 0$. Therefore, we can write $(\overrightarrow r - \overrightarrow c)$ as a scalar multiple of $\overrightarrow b$. Let this scalar be $\lambda$. $\overrightarrow r - \overrightarrow c = \lambda \overrightarrow b$ Rearranging this equation to express $\overrightarrow r$ in terms of $\overrightarrow b$, $\overrightarrow c$, and $\lambda$: $\overrightarrow r = \overrightarrow c + \lambda \overrightarrow b \quad (*)$ This gives us a general form for $\overrightarrow r$.

Step 2: Use the dot product condition to determine the scalar $\lambda$.

The second given condition is $\overrightarrow r \,.\,\overrightarrow a = 0$. Substitute the expression for $\overrightarrow r$ from Step 1 into this equation: $(\overrightarrow c + \lambda \overrightarrow b) \,.\,\overrightarrow a = 0$ Using the distributive property of the dot product: $\overrightarrow c \,.\,\overrightarrow a + (\lambda \overrightarrow b) \,.\,\overrightarrow a = 0$ Using the scalar multiplication property of the dot product: $\overrightarrow c \,.\,\overrightarrow a + \lambda (\overrightarrow b \,.\,\overrightarrow a) = 0 \quad (**)$ Now, we need to calculate the dot products $\overrightarrow c \,.\,\overrightarrow a$ and $\overrightarrow b \,.\,\overrightarrow a$. $\overrightarrow a = 1\widehat i + 0\widehat j + 2\widehat k$ $\overrightarrow b = 1\widehat i + 1\widehat j + 1\widehat k$ $\overrightarrow c = 7\widehat i - 3\widehat j + 4\widehat k$

Calculating $\overrightarrow c \,.\,\overrightarrow a$: $\overrightarrow c \,.\,\overrightarrow a = (7)(1) + (-3)(0) + (4)(2) = 7 + 0 + 8 = 15$ Calculating $\overrightarrow b \,.\,\overrightarrow a$: $\overrightarrow b \,.\,\overrightarrow a = (1)(1) + (1)(0) + (1)(2) = 1 + 0 + 2 = 3$ Substitute these values back into equation $(**)$: $15 + \lambda (3) = 0$ $3\lambda = -15$ $\lambda = -5$ We have now found the specific value of the scalar $\lambda$.

Step 3: Calculate the required dot product $\overrightarrow r \,.\,\overrightarrow c$.

We need to find the value of $\overrightarrow r \,.\,\overrightarrow c$. Substitute the expression for $\overrightarrow r$ from Step 1 ($\overrightarrow r = \overrightarrow c + \lambda \overrightarrow b$) and the found value of $\lambda = -5$ into the expression we need to calculate: $\overrightarrow r \,.\,\overrightarrow c = (\overrightarrow c + \lambda \overrightarrow b) \,.\,\overrightarrow c$ Using the distributive property of the dot product: $\overrightarrow r \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow c + \lambda (\overrightarrow b \,.\,\overrightarrow c)$ Now, we need to calculate the dot products $\overrightarrow c \,.\,\overrightarrow c$ and $\overrightarrow b \,.\,\overrightarrow c$.

Calculating $\overrightarrow c \,.\,\overrightarrow c$: $\overrightarrow c \,.\,\overrightarrow c = (7)(7) + (-3)(-3) + (4)(4) = 49 + 9 + 16 = 74$ Calculating $\overrightarrow b \,.\,\overrightarrow c$: $\overrightarrow b \,.\,\overrightarrow c = (1)(7) + (1)(-3) + (1)(4) = 7 - 3 + 4 = 8$ Substitute these values and $\lambda = -5$ into the equation for $\overrightarrow r \,.\,\overrightarrow c$: $\overrightarrow r \,.\,\overrightarrow c = 74 + (-5)(8)$ $\overrightarrow r \,.\,\overrightarrow c = 74 - 40$ $\overrightarrow r \,.\,\overrightarrow c = 34$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs, especially when using the anticommutativity of the cross product.
• Component Calculation: When calculating dot products, ensure you multiply corresponding components and then sum them up correctly.
• Algebraic Manipulation: Ensure all algebraic steps, particularly when solving for $\lambda$, are performed accurately.

Summary

The problem involves finding an unknown vector $\overrightarrow r$ that satisfies two conditions: one related to the cross product and another to the dot product. We first simplify the cross product condition to express $\overrightarrow r$ in a general form involving a scalar parameter $\lambda$. Then, we use the dot product condition to form an equation that allows us to solve for $\lambda$. Finally, we substitute the found value of $\lambda$ back into the general form of $\overrightarrow r$ to calculate the required dot product $\overrightarrow r \,.\,\overrightarrow c$.

The final answer is $\boxed{\text{34}}$.