Key Concepts and Formulas

• Linear Independence of Vectors: A set of vectors $\{\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n\}$ is linearly independent if the only solution to the equation $c_1\vec{v}_1 + c_2\vec{v}_2 + \dots + c_n\vec{v}_n = \vec{0}$ is $c_1 = c_2 = \dots = c_n = 0$. The problem statement directly gives this condition for vectors $\vec{a}, \vec{b}, \vec{c}$.
• Scalar Triple Product: For three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ in 3D space, their scalar triple product is given by $\vec{a} \cdot (\vec{b} \times \vec{c})$. This value represents the signed volume of the parallelepiped formed by the three vectors.
• Geometric Interpretation of Scalar Triple Product: Three vectors are linearly independent (i.e., they are not coplanar) if and only if their scalar triple product is non-zero. The scalar triple product can be calculated as the determinant of the matrix formed by the components of the vectors.

Step-by-Step Solution

Step 1: Understand the Condition for Linear Independence The problem states that for $\alpha, \beta, \gamma \in \mathbf{R}$, the equation $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0}$ implies $\alpha=\beta=\gamma=0$. This is the definition of linear independence for the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$. In three-dimensional space, three vectors are linearly independent if and only if they are non-coplanar.

Step 2: Relate Linear Independence to the Scalar Triple Product For three vectors in 3D space, linear independence is equivalent to them being non-coplanar. Non-coplanar vectors form a parallelepiped with a non-zero volume. The volume of the parallelepiped formed by vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by the absolute value of their scalar triple product, $[\vec{a} \vec{b} \vec{c}]$. Therefore, for $\vec{a}$, $\vec{b}$, and $\vec{c}$ to be linearly independent, their scalar triple product must be non-zero: $[\vec{a} \vec{b} \vec{c}] \ne 0$.

Step 3: Write Down the Components of the Vectors The given vectors are: $\vec{a} = (1+t) \hat{i} + (1-t) \hat{j} + 1 \hat{k}$ $\vec{b} = (1-t) \hat{i} + (1+t) \hat{j} + 2 \hat{k}$ $\vec{c} = t \hat{i} - t \hat{j} + 1 \hat{k}$

Step 4: Set Up the Determinant for the Scalar Triple Product The scalar triple product $[\vec{a} \vec{b} \vec{c}]$ can be computed as the determinant of the matrix whose rows (or columns) are the components of the vectors: $[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1 \end{vmatrix}$ For linear independence, this determinant must be non-zero: $\begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1 \end{vmatrix} \ne 0$

Step 5: Expand the Determinant We expand the determinant. Let's expand along the first row: $(1+t) \begin{vmatrix} 1+t & 2 \\ -t & 1 \end{vmatrix} - (1-t) \begin{vmatrix} 1-t & 2 \\ t & 1 \end{vmatrix} + 1 \begin{vmatrix} 1-t & 1+t \\ t & -t \end{vmatrix} \ne 0$

Step 6: Evaluate the $2 \times 2$ Determinants Now, we calculate each $2 \times 2$ determinant:

• $\begin{vmatrix} 1+t & 2 \\ -t & 1 \end{vmatrix} = (1+t)(1) - (2)(-t) = 1+t+2t = 1+3t$
• $\begin{vmatrix} 1-t & 2 \\ t & 1 \end{vmatrix} = (1-t)(1) - (2)(t) = 1-t-2t = 1-3t$
• $\begin{vmatrix} 1-t & 1+t \\ t & -t \end{vmatrix} = (1-t)(-t) - (1+t)(t) = (-t+t^2) - (t+t^2) = -t+t^2-t-t^2 = -2t$

Step 7: Substitute and Simplify the Expression Substitute these values back into the expanded determinant: $(1+t)(1+3t) - (1-t)(1-3t) + 1(-2t) \ne 0$ Expand the products:

• $(1+t)(1+3t) = 1 + 3t + t + 3t^2 = 3t^2 + 4t + 1$
• $(1-t)(1-3t) = 1 - 3t - t + 3t^2 = 3t^2 - 4t + 1$

Now, substitute these back into the inequality: $(3t^2 + 4t + 1) - (3t^2 - 4t + 1) - 2t \ne 0$ Remove the parentheses, being careful with the signs: $3t^2 + 4t + 1 - 3t^2 + 4t - 1 - 2t \ne 0$ Combine like terms: $(3t^2 - 3t^2) + (4t + 4t - 2t) + (1 - 1) \ne 0$ $0 + 6t + 0 \ne 0$ $6t \ne 0$

Step 8: Determine the Set of Values for $t$ The inequality $6t \ne 0$ simplifies to $t \ne 0$. This means that $t$ can be any real number except for $0$. The set of all such values of $t$ is $\mathbf{R} - \{0\}$.

Common Mistakes & Tips

• Algebraic Errors: Be meticulous when expanding determinants and simplifying algebraic expressions. A single sign error can lead to an incorrect result. Double-check your calculations, especially when dealing with subtractions.
• Misinterpreting Linear Independence: Confusing linear independence with linear dependence can lead to setting the determinant to zero instead of non-zero. Remember that linear independence means the vectors are not coplanar, hence their volume (scalar triple product) is non-zero.
• Determinant Expansion: Ensure correct application of the cofactor expansion for determinants. For a $3 \times 3$ determinant, the signs alternate: +, -, +.

Summary

The condition that $\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0}$ implies $\alpha=\beta=\gamma=0$ is the definition of linear independence for the vectors $\vec{a}, \vec{b}, \vec{c}$. In 3D space, three vectors are linearly independent if and only if they are non-coplanar, which means their scalar triple product must be non-zero. We calculated the scalar triple product as a determinant and found that it simplifies to $6t$. For the vectors to be linearly independent, $6t \ne 0$, which implies $t \ne 0$. Therefore, the set of all possible values for $t$ is all real numbers except zero, denoted as $\mathbf{R} - \{0\}$.

The final answer is \boxed{\text{C}}.