1. Key Concepts and Formulas

• Magnitude of a Vector: For any vector $\vec{v}$, its magnitude is denoted by $|\vec{v}|$. The square of the magnitude is $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.
• Dot Product Properties:
  • $\vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x}$ (Commutative property)
  • $\vec{x} \cdot (\vec{y} + \vec{z}) = \vec{x} \cdot \vec{y} + \vec{x} \cdot \vec{z}$ (Distributive property)
  • $(\vec{x} + \vec{y}) \cdot (\vec{x} + \vec{y}) = |\vec{x}|^2 + 2(\vec{x} \cdot \vec{y}) + |\vec{y}|^2$
  • If $\vec{x} \cdot \vec{y} = 0$ and $\vec{x}, \vec{y}$ are nonzero, then $\vec{x}$ and $\vec{y}$ are perpendicular.
• Vector Addition and Subtraction: The sum and difference of vectors follow standard algebraic rules.

2. Step-by-Step Solution

We are given $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, and that $\vec{b}$ and $\vec{c}$ are nonzero vectors satisfying $|\vec{a}+\vec{b}+\vec{c}| = |\vec{a}+\vec{b}-\vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. We need to evaluate two statements: (A) $|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$ for all $\lambda \in \mathbb{R}$, and (B) $\vec{a}$ and $\vec{c}$ are always parallel.

Step 1: Simplify the given magnitude equality. We are given $|\vec{a}+\vec{b}+\vec{c}| = |\vec{a}+\vec{b}-\vec{c}|$. Squaring both sides, we get: $|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}+\vec{b}-\vec{c}|^2$ Using the property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$: $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = (\vec{a}+\vec{b}-\vec{c}) \cdot (\vec{a}+\vec{b}-\vec{c})$ Expand both sides: $(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) + 2(\vec{a}+\vec{b}) \cdot \vec{c} + \vec{c} \cdot \vec{c} = (\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) - 2(\vec{a}+\vec{b}) \cdot \vec{c} + \vec{c} \cdot \vec{c}$ $|\vec{a}+\vec{b}|^2 + 2(\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}) + |\vec{c}|^2 = |\vec{a}+\vec{b}|^2 - 2(\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}) + |\vec{c}|^2$ Subtract $|\vec{a}+\vec{b}|^2 + |\vec{c}|^2$ from both sides: $2(\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}) = -2(\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c})$ $4(\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c}) = 0$ $\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0$

Step 2: Use the given condition $\vec{b} \cdot \vec{c} = 0$. Substitute $\vec{b} \cdot \vec{c} = 0$ into the equation from Step 1: $\vec{a} \cdot \vec{c} + 0 = 0$ $\vec{a} \cdot \vec{c} = 0$ This implies that vector $\vec{a}$ is perpendicular to vector $\vec{c}$, provided that $\vec{a}$ and $\vec{c}$ are nonzero. We are given that $\vec{b}$ and $\vec{c}$ are nonzero. We also know $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, which is a nonzero vector.

Step 3: Evaluate Statement (B): $\vec{a}$ and $\vec{c}$ are always parallel. From Step 2, we found that $\vec{a} \cdot \vec{c} = 0$. For two nonzero vectors to be parallel, their dot product must be non-zero in a way that reflects their alignment (e.g., $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos(0)$ or $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}| \cos(\pi)$). The condition $\vec{a} \cdot \vec{c} = 0$ implies that $\vec{a}$ and $\vec{c}$ are perpendicular (since both are nonzero). Therefore, $\vec{a}$ and $\vec{c}$ are not always parallel. Statement (B) is incorrect.

Step 4: Evaluate Statement (A): $|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$ for all $\lambda \in \mathbb{R}$. We need to check if $|\vec{a}+\lambda \vec{c}|^2 \geq |\vec{a}|^2$ for all $\lambda \in \mathbb{R}$. $|\vec{a}+\lambda \vec{c}|^2 = (\vec{a}+\lambda \vec{c}) \cdot (\vec{a}+\lambda \vec{c})$ $= \vec{a} \cdot \vec{a} + 2(\vec{a} \cdot (\lambda \vec{c})) + (\lambda \vec{c}) \cdot (\lambda \vec{c})$ $= |\vec{a}|^2 + 2\lambda (\vec{a} \cdot \vec{c}) + \lambda^2 |\vec{c}|^2$ From Step 2, we know that $\vec{a} \cdot \vec{c} = 0$. Substituting this into the equation: $|\vec{a}+\lambda \vec{c}|^2 = |\vec{a}|^2 + 2\lambda (0) + \lambda^2 |\vec{c}|^2$ $|\vec{a}+\lambda \vec{c}|^2 = |\vec{a}|^2 + \lambda^2 |\vec{c}|^2$ We need to check if $|\vec{a}|^2 + \lambda^2 |\vec{c}|^2 \geq |\vec{a}|^2$ for all $\lambda \in \mathbb{R}$. This simplifies to: $\lambda^2 |\vec{c}|^2 \geq 0$ Since $\vec{c}$ is a nonzero vector, $|\vec{c}|^2 > 0$. Also, $\lambda^2 \geq 0$ for all real numbers $\lambda$. Therefore, $\lambda^2 |\vec{c}|^2 \geq 0$ is always true for all $\lambda \in \mathbb{R}$. This means $|\vec{a}+\lambda \vec{c}|^2 \geq |\vec{a}|^2$ for all $\lambda \in \mathbb{R}$, which implies $|\vec{a}+\lambda \vec{c}| \geq |\vec{a}|$ for all $\lambda \in \mathbb{R}$. Statement (A) is correct.

3. Common Mistakes & Tips

• Algebraic Errors: Be very careful when expanding dot products of sums of vectors. A small mistake can lead to an incorrect conclusion.
• Misinterpreting Dot Product: Remember that a zero dot product between two nonzero vectors means they are perpendicular, not parallel.
• Squared Magnitudes: Always square both sides of an equation involving magnitudes before expanding, as it simplifies nicely using the dot product property $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.

4. Summary

The problem starts by simplifying the given equality of vector magnitudes. Squaring both sides and using the properties of the dot product, along with the given condition $\vec{b} \cdot \vec{c} = 0$, leads to the crucial result that $\vec{a} \cdot \vec{c} = 0$. This indicates that $\vec{a}$ and $\vec{c}$ are perpendicular. Consequently, statement (B) that $\vec{a}$ and $\vec{c}$ are always parallel is false. For statement (A), we consider the square of the magnitude of $\vec{a}+\lambda \vec{c}$ and use the fact that $\vec{a} \cdot \vec{c} = 0$. This simplifies to $|\vec{a}+\lambda \vec{c}|^2 = |\vec{a}|^2 + \lambda^2 |\vec{c}|^2$. Since $\lambda^2 |\vec{c}|^2 \geq 0$, the inequality $|\vec{a}+\lambda \vec{c}| \geq|\vec{a}|$ holds true for all real values of $\lambda$. Therefore, only statement (A) is correct.

5. Final Answer

The final answer is \boxed{A}