Key Concepts and Formulas

• Vector in a Plane: A vector $\vec{v}$ lying in the plane of two non-collinear vectors $\vec{a}$ and $\vec{b}$ can be represented as a linear combination: $\vec{v} = \lambda \vec{a} + \mu \vec{b}$, where $\lambda$ and $\mu$ are scalars.
• Perpendicular Vectors: Two vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if their dot product is zero: $\vec{u} \cdot \vec{v} = 0$.
• Unit Vector: A unit vector $\hat{u}$ in the direction of a non-zero vector $\vec{u}$ is given by $\hat{u} = \frac{\vec{u}}{|\vec{u}|}$, where $|\vec{u}|$ is the magnitude of $\vec{u}$.

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Step-by-Step Solution

Step 1: Express the vector in the plane We are looking for a unit vector $\hat{c}$ that lies in the plane of $\vec{a}$ and $\vec{b}$. Therefore, we can express $\hat{c}$ (or a vector proportional to it) as a linear combination of $\vec{a}$ and $\vec{b}$. Let this vector be $\vec{v}$. $\vec{v} = \lambda \vec{a} + \mu \vec{b}$ where $\lambda$ and $\mu$ are scalar constants.

Step 2: Apply the perpendicularity condition We are given that $\hat{c}$ is perpendicular to $\vec{a}$. Since $\vec{v}$ is in the same direction as $\hat{c}$, $\vec{v}$ must also be perpendicular to $\vec{a}$. This means their dot product is zero. $\vec{v} \cdot \vec{a} = 0$ Substituting the expression for $\vec{v}$: $(\lambda \vec{a} + \mu \vec{b}) \cdot \vec{a} = 0$ Using the distributive property of the dot product: $\lambda (\vec{a} \cdot \vec{a}) + \mu (\vec{b} \cdot \vec{a}) = 0$

Step 3: Calculate the required dot products We are given $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Let's calculate $\vec{a} \cdot \vec{a}$ and $\vec{b} \cdot \vec{a}$. $\vec{a} \cdot \vec{a} = (1)(1) + (2)(2) + (1)(1) = 1 + 4 + 1 = 6$ (Note: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$) $\vec{b} \cdot \vec{a} = (2)(1) + (1)(2) + (-1)(1) = 2 + 2 - 1 = 3$

Step 4: Solve for the relationship between $\lambda$ and $\mu$ Substitute the calculated dot products back into the equation from Step 2: $\lambda (6) + \mu (3) = 0$ Dividing the equation by 3, we get: $2\lambda + \mu = 0$ This gives us the relationship: $\mu = -2\lambda$

Step 5: Determine the direction vector $\vec{v}$ Substitute the relationship $\mu = -2\lambda$ back into the expression for $\vec{v}$ from Step 1: $\vec{v} = \lambda \vec{a} + (-2\lambda) \vec{b}$ Factor out $\lambda$: $\vec{v} = \lambda (\vec{a} - 2\vec{b})$ Now, let's compute the vector $(\vec{a} - 2\vec{b})$: $\vec{a} - 2\vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) - 2(2\hat{i} + \hat{j} - \hat{k})$ $\vec{a} - 2\vec{b} = (\hat{i} + 2\hat{j} + \hat{k}) - (4\hat{i} + 2\hat{j} - 2\hat{k})$ $\vec{a} - 2\vec{b} = (1-4)\hat{i} + (2-2)\hat{j} + (1-(-2))\hat{k}$ $\vec{a} - 2\vec{b} = -3\hat{i} + 0\hat{j} + 3\hat{k}$ $\vec{a} - 2\vec{b} = -3\hat{i} + 3\hat{k}$ So, $\vec{v} = \lambda (-3\hat{i} + 3\hat{k})$. Since $\lambda$ is a non-zero scalar, the direction of $\vec{v}$ is the same as the direction of $-3\hat{i} + 3\hat{k}$. We can simplify this direction vector by dividing by -3 to get $-\hat{i} + \hat{k}$. Let's use $\vec{d} = -\hat{i} + \hat{k}$ as our direction vector.

Step 6: Normalize the direction vector to find the unit vector $\hat{c}$ The problem asks for a unit vector. We need to find the magnitude of $\vec{d} = -\hat{i} + \hat{k}$. $|\vec{d}| = |-\hat{i} + \hat{k}| = \sqrt{(-1)^2 + (0)^2 + (1)^2} = \sqrt{1 + 0 + 1} = \sqrt{2}$ The unit vector $\hat{c}$ is obtained by dividing $\vec{d}$ by its magnitude: $\hat{c} = \frac{\vec{d}}{|\vec{d}|} = \frac{-\hat{i} + \hat{k}}{\sqrt{2}} = \frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$

Step 7: Compare with the given options The calculated unit vector $\hat{c} = \frac{1}{\sqrt{2}}(-\hat{i} + \hat{k})$ matches option (A).

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Common Mistakes & Tips

• Misinterpreting "in the plane": Ensure you use the linear combination $\lambda \vec{a} + \mu \vec{b}$ correctly.
• Forgetting to normalize: The question specifically asks for a unit vector. Always normalize your result if needed.
• Arithmetic errors: Double-check dot product calculations and vector subtractions, as these are common sources of mistakes.

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Summary

The problem requires finding a unit vector that lies in the plane of two given vectors and is perpendicular to one of them. We achieved this by representing the vector as a linear combination of the given vectors, using the perpendicularity condition to establish a relationship between the scalar coefficients, and then normalizing the resulting direction vector. The key steps involved understanding vector representation in a plane and applying the dot product property for perpendicularity.

The final answer is $\boxed{\text{(A)}}$.