Key Concepts and Formulas

1. Vector in a Plane: If a vector $\overrightarrow v$ lies in the plane formed by two non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$, it can be uniquely expressed as a linear combination of these two vectors: $\overrightarrow v = \lambda_1 \overrightarrow a + \lambda_2 \overrightarrow b$ where $\lambda_1$ and $\lambda_2$ are scalar constants. This is fundamental for defining an unknown vector within a specific geometric constraint.

2. Scalar Projection of a Vector: The scalar projection of vector $\overrightarrow v$ onto vector $\overrightarrow c$ measures the "shadow" of $\overrightarrow v$ cast along the direction of $\overrightarrow c$. It is given by the formula: $\text{Proj}_{\overrightarrow c} \overrightarrow v = \frac{\overrightarrow v \cdot \overrightarrow c}{|\overrightarrow c|}$ Here, $\overrightarrow v \cdot \overrightarrow c$ is the dot product of $\overrightarrow v$ and $\overrightarrow c$, and $|\overrightarrow c|$ is the magnitude of $\overrightarrow c$.

3. Dot Product Properties:

   • For two vectors $\overrightarrow A = A_x \widehat i + A_y \widehat j + A_z \widehat k$ and $\overrightarrow B = B_x \widehat i + B_y \widehat j + B_z \widehat k$, their dot product is: $\overrightarrow A \cdot \overrightarrow B = A_x B_x + A_y B_y + A_z B_z$
   • The dot product of a vector $\overrightarrow v$ with a unit vector along an axis gives the component of $\overrightarrow v$ along that axis. For example, $\overrightarrow v \cdot \widehat j = v_y$ (the $\widehat j$-component of $\overrightarrow v$).

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Step-by-Step Solution

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

Our goal is to find $\overrightarrow v \cdot \left( {\widehat i + \widehat k} \right)$.

Step 1: Expressing $\overrightarrow v$ as a linear combination of $\overrightarrow a$ and $\overrightarrow b$.

Why this step? The problem states that $\overrightarrow v$ lies in the plane of $\overrightarrow a$ and $\overrightarrow b$. This geometric constraint allows us to represent $\overrightarrow v$ algebraically as a linear combination of $\overrightarrow a$ and $\overrightarrow b$ using two unknown scalar coefficients, $\lambda_1$ and $\lambda_2$.

Let $\overrightarrow v = \lambda_1 \overrightarrow a + \lambda_2 \overrightarrow b$. Substitute the given expressions for $\overrightarrow a$ and $\overrightarrow b$: $\overrightarrow v = \lambda_1 (\widehat i + \widehat j + 2\widehat k) + \lambda_2 (2\widehat i - 3\widehat j + \widehat k)$ Now, we group the components for $\widehat i$, $\widehat j$, and $\widehat k$: $\overrightarrow v = (\lambda_1 + 2\lambda_2)\widehat i + (\lambda_1 - 3\lambda_2)\widehat j + (2\lambda_1 + \lambda_2)\widehat k \quad (*)$ This expression for $\overrightarrow v$ is crucial as we will use the given conditions to solve for $\lambda_1$ and $\lambda_2$.

Step 2: Utilizing the Projection Condition.

Why this step? We are given that the scalar projection of $\overrightarrow v$ on $\overrightarrow c$ is $\frac{2}{\sqrt{3}}$. This will provide our first equation involving $\lambda_1$ and $\lambda_2$. First, we need to calculate the magnitude of $\overrightarrow c$ and the dot product $\overrightarrow v \cdot \overrightarrow c$.

First, calculate the magnitude of $\overrightarrow c$: $|\overrightarrow c| = |\widehat i - \widehat j + \widehat k| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$

Next, calculate the dot product $\overrightarrow v \cdot \overrightarrow c$ using the component form of $\overrightarrow v$ from $(*)$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$: $\overrightarrow v \cdot \overrightarrow c = \left( (\lambda_1 + 2\lambda_2)\widehat i + (\lambda_1 - 3\lambda_2)\widehat j + (2\lambda_1 + \lambda_2)\widehat k \right) \cdot (\widehat i - \widehat j + \widehat k)$ Applying the dot product formula: $= (\lambda_1 + 2\lambda_2)(1) + (\lambda_1 - 3\lambda_2)(-1) + (2\lambda_1 + \lambda_2)(1)$ $= \lambda_1 + 2\lambda_2 - \lambda_1 + 3\lambda_2 + 2\lambda_1 + \lambda_2$ Combine the terms for $\lambda_1$ and $\lambda_2$: $= (1 - 1 + 2)\lambda_1 + (2 + 3 + 1)\lambda_2$ $= 2\lambda_1 + 6\lambda_2$

Now, apply the scalar projection formula and set it equal to the given value: $\text{Proj}_{\overrightarrow c} \overrightarrow v = \frac{\overrightarrow v \cdot \overrightarrow c}{|\overrightarrow c|} = \frac{2\lambda_1 + 6\lambda_2}{\sqrt{3}}$ We are given that this projection is $\frac{2}{\sqrt{3}}$. So: $\frac{2\lambda_1 + 6\lambda_2}{\sqrt{3}} = \frac{2}{\sqrt{3}}$ Multiply both sides by $\sqrt{3}$: $2\lambda_1 + 6\lambda_2 = 2$ Divide by 2 to simplify: $\lambda_1 + 3\lambda_2 = 1 \quad \text{(Equation 1)}$

Step 3: Applying the Dot Product Condition.

Why this step? We are given the condition $\overrightarrow v \cdot \widehat j = 7$. This is a direct way to find the $\widehat j$-component of $\overrightarrow v$, which provides our second equation for $\lambda_1$ and $\lambda_2$.

From the component form of $\overrightarrow v$ in $(*)$, the $\widehat j$-component of $\overrightarrow v$ is $(\lambda_1 - 3\lambda_2)$. Therefore, we can directly equate this component to 7: $\lambda_1 - 3\lambda_2 = 7 \quad \text{(Equation 2)}$

Step 4: Solving for the Scalar Constants $\lambda_1$ and $\lambda_2$.

Why this step? We now have a system of two linear equations with two unknowns. Solving this system will give us the specific values of $\lambda_1$ and $\lambda_2$, which are essential to fully define $\overrightarrow v$.

Our system of equations is:

1. $\lambda_1 + 3\lambda_2 = 1$
2. $\lambda_1 - 3\lambda_2 = 7$

To solve, we can add Equation 1 and Equation 2. This will eliminate $\lambda_2$: $(\lambda_1 + 3\lambda_2) + (\lambda_1 - 3\lambda_2) = 1 + 7$ $2\lambda_1 = 8$ $\lambda_1 = 4$

Now, substitute $\lambda_1 = 4$ into Equation 1: $4 + 3\lambda_2 = 1$ $3\lambda_2 = 1 - 4$ $3\lambda_2 = -3$ $\lambda_2 = -1$

So, we have found the scalar constants: $\lambda_1 = 4$ and $\lambda_2 = -1$.

Step 5: Determining the Vector $\overrightarrow v$.

Why this step? With the values of $\lambda_1$ and $\lambda_2$ determined, we can substitute them back into the general expression for $\overrightarrow v$ from Step 1 to find its explicit vector form.

Substitute $\lambda_1 = 4$ and $\lambda_2 = -1$ into $\overrightarrow v = (\lambda_1 + 2\lambda_2)\widehat i + (\lambda_1 - 3\lambda_2)\widehat j + (2\lambda_1 + \lambda_2)\widehat k$: $\overrightarrow v = (4 + 2(-1))\widehat i + (4 - 3(-1))\widehat j + (2(4) + (-1))\widehat k$ $\overrightarrow v = (4 - 2)\widehat i + (4 + 3)\widehat j + (8 - 1)\widehat k$ $\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k$

Step 6: Calculating the Final Desired Dot Product.

Why this step? The problem asks for the value of $\overrightarrow v \cdot (\widehat i + \widehat k)$. Now that we have fully determined $\overrightarrow v$, we can directly compute this dot product.

Using $\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k$: $\overrightarrow v \cdot (\widehat i + \widehat k) = (2\widehat i + 7\widehat j + 7\widehat k) \cdot (1\widehat i + 0\widehat j + 1\widehat k)$ Applying the dot product formula: $= (2)(1) + (7)(0) + (7)(1)$ $= 2 + 0 + 7$ $= 9$

Thus, $\overrightarrow v \cdot (\widehat i + \widehat k) = 9$.

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Common Mistakes & Tips

• Systematic Approach: Break down the problem into smaller, manageable steps. Start by defining the unknown vector, then use each given condition to form an equation.
• Careful Algebra: Errors in combining terms or solving the system of equations are common. Double-check your arithmetic, especially when dealing with negative signs.
• Understanding Vector in a Plane: Remember that any vector in the plane of two non-collinear vectors $\overrightarrow a$ and $\overrightarrow b$ can always be written as $\lambda_1 \overrightarrow a + \lambda_2 \overrightarrow b$. This is a crucial starting point for many vector problems.
• Projection Formula: Ensure you correctly use the scalar projection formula, $\frac{\overrightarrow v \cdot \overrightarrow c}{|\overrightarrow c|}$, and not the vector projection formula, $\frac{(\overrightarrow v \cdot \overrightarrow c)}{|\overrightarrow c|^2}\overrightarrow c$.
• Dot Product Interpretation: Recognize that $\overrightarrow v \cdot \widehat j$ directly yields the $\widehat j$-component of $\overrightarrow v$. This shortcut can save time.

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Summary

This problem demonstrates a common strategy for solving vector problems involving unknown vectors subject to various conditions. We first represented the unknown vector $\overrightarrow v$ as a linear combination of the given vectors $\overrightarrow a$ and $\overrightarrow b$. Then, we translated each given condition (scalar projection onto $\overrightarrow c$ and dot product with $\widehat j$) into algebraic equations involving the scalar coefficients $\lambda_1$ and $\lambda_2$. Solving this system of equations yielded $\lambda_1 = 4$ and $\lambda_2 = -1$. Substituting these values back into the expression for $\overrightarrow v$ gave $\overrightarrow v = 2\widehat i + 7\widehat j + 7\widehat k$. Finally, we computed the desired dot product $\overrightarrow v \cdot (\widehat i + \widehat k)$, which resulted in $9$.

The final answer is $\boxed{9}$.

Note: Based on the provided problem statement and standard vector mathematics, the derived answer is 9, which corresponds to option (D). If the intended answer was 6 (option A), there might be a subtle difference in the problem statement or the provided options, as the conditions given consistently lead to 9.