Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar if they lie in the same plane. This means that one of the vectors can be expressed as a linear combination of the other two, or equivalently, their scalar triple product is zero.
• Scalar Triple Product: The scalar triple product of three vectors $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$, $\overrightarrow b = b_1\widehat i + b_2\widehat j + b_3\widehat k$, and $\overrightarrow c = c_1\widehat i + c_2\widehat j + c_3\widehat k$ is given by the determinant of the matrix formed by their components: $[\overrightarrow a \, \overrightarrow b \, \overrightarrow c] = \overrightarrow a \cdot (\overrightarrow b \times \overrightarrow c) = \left| {\begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{matrix} } \right|$
• Condition for Coplanarity: Three vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar if and only if their scalar triple product is zero: $[\overrightarrow a \, \overrightarrow b \, \overrightarrow c] = 0$.

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Step-by-Step Solution

Step 1: Understand the Given Information and the Problem We are given three vectors: $\overrightarrow a = \widehat i + \widehat j + \widehat k$ $\overrightarrow b = \widehat i - \widehat j + 2\widehat k$ $\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k$

We are told that vector $\overrightarrow c$ lies in the plane of vectors $\overrightarrow a$ and $\overrightarrow b$. This implies that the three vectors $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$ are coplanar.

Step 2: Apply the Coplanarity Condition For three vectors to be coplanar, their scalar triple product must be equal to zero. Therefore, we have: $[\overrightarrow a \, \overrightarrow b \, \overrightarrow c] = 0$

Step 3: Formulate the Determinant for the Scalar Triple Product The components of the given vectors are: For $\overrightarrow a$: $(1, 1, 1)$ For $\overrightarrow b$: $(1, -1, 2)$ For $\overrightarrow c$: $(x, x-2, -1)$

We can set up the determinant using these components: $\left| {\begin{matrix} 1 & 1 & 1 \\ 1 & { - 1} & 2 \\ x & {\left( {x - 2} \right)} & { - 1} \\ \end{matrix} } \right| = 0$

Step 4: Evaluate the Determinant We will expand the determinant along the first row. The expansion of a $3 \times 3$ determinant $\left| {\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} } \right|$ is $a(ei - fh) - b(di - fg) + c(dh - eg)$.

Applying this to our determinant: $1 \cdot \left( {(-1)(-1) - (2)(x-2)} \right) - 1 \cdot \left( {(1)(-1) - (2)(x)} \right) + 1 \cdot \left( {(1)(x-2) - (-1)(x)} \right) = 0$

Step 5: Simplify the Algebraic Expression Let's simplify each term within the determinant expansion:

• First term: $1 \cdot (1 - (2x - 4)) = 1 - 2x + 4 = 5 - 2x$
• Second term: $-1 \cdot (-1 - 2x) = 1 + 2x$
• Third term: $1 \cdot (x - 2 + x) = 2x - 2$

Now, substitute these simplified terms back into the equation: $(5 - 2x) + (1 + 2x) + (2x - 2) = 0$

Step 6: Solve the Linear Equation for x Combine the constant terms and the terms involving $x$: $(5 + 1 - 2) + (-2x + 2x + 2x) = 0$ $4 + 2x = 0$

Now, isolate $x$: $2x = -4$ $x = \frac{-4}{2}$ $x = -2$

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Common Mistakes & Tips

• Sign Errors in Determinant Expansion: Be extremely careful with the alternating signs ($+ - +$) when expanding the determinant. A small sign error can lead to an incorrect final answer.
• Algebraic Simplification: Ensure accurate simplification of expressions involving parentheses and combining like terms. Mistakes in algebra are common and can be costly.
• Understanding Coplanarity: The core concept is that coplanar vectors have a zero scalar triple product. If the problem states vectors are coplanar, set the determinant to zero.

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Summary

The problem states that vector $\overrightarrow c$ lies in the plane of vectors $\overrightarrow a$ and $\overrightarrow b$, which means the three vectors are coplanar. The condition for coplanarity is that their scalar triple product must be zero. By forming a determinant with the components of the vectors and setting it to zero, we derived a linear equation in $x$. Solving this equation, we found that $x = -2$.

The final answer is $\boxed{-2}$.