Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Let $\vec{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\vec{c}$ is:

Step 1: Express vector c in the plane of a and b

Since vector $\vec{c}$ lies in the plane of vectors $\vec{a}$ and $\vec{b}$, it can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$. Here, $x$ and $y$ are scalar coefficients.

Step 2: Use the perpendicularity condition

We are given that $\vec{c}$ is perpendicular to $\vec{a}$. This means their dot product is zero. Substituting the expression for $\vec{c}$, we get an equation involving $x$ and $y$. First, let's calculate the dot products $\vec{a} \cdot \vec{a}$ and $\vec{a} \cdot \vec{b}$.

Step 3: Calculate dot products and solve for x/y

We calculate $\vec{a} \cdot \vec{a} = (1)(1) + (2)(2) + (1)(1) = 1 + 4 + 1 = 6$. And $\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (1)(-1) = 2 + 2 - 1 = 3$. Substituting these into the dot product equation, we get $6x + 3y = 0$, which simplifies to $y = -2x$. (Correction: $\vec{a} \cdot \vec{b} = 3$, so $6x+3y=0 \implies y=-2x$). Let's recheck the calculation. $\vec{a} \cdot \vec{b} = (1)(2) + (2)(1) + (1)(-1) = 2 + 2 - 1 = 3$. So $6x + 3y = 0 \implies y = -2x$.

Step 4: Substitute y and find vector c

Now we substitute $y = -2x$ back into the expression for $\vec{c}$. This gives us $\vec{c} = x\vec{a} - 2x\vec{b} = x(\vec{a} - 2\vec{b})$. Let's calculate the vector $\vec{a} - 2\vec{b}$.

Step 5: Calculate a - 2b and its magnitude

We compute $\vec{a} - 2\vec{b} = (1 - 4)\hat{i} + (2 - 2)\hat{j} + (1 - (-2))\hat{k} = -3\hat{i} + 0\hat{j} + 3\hat{k} = -3\hat{i} + 3\hat{k}$. The magnitude of this vector is $\sqrt{(-3)^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.

Step 6: Normalize vector c

Since $\vec{c}$ is a unit vector, its magnitude is 1. We have $|\vec{c}| = |x| |\vec{a} - 2\vec{b}|$. Setting this to 1, we find the possible values for $x$.

Step 7: Final vector c

Substitute the value of $x$ back into the expression for $\vec{c}$. This gives us the two possible unit vectors that satisfy the given conditions. We can simplify the expression by factoring out 3 from the vector and canceling it with the 3 in the denominator.

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