If $\vec{a}$ is a nonzero vector such that its projections on the vectors $2\hat{i}-\hat{j}+2\hat{k}$, $\hat{i}+2\hat{j}-2\hat{k}$, and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is:

Step 1: Define the given vectors and projection formula

We are given three vectors, let's call them $\vec{b}_{1}$, $\vec{b}_{2}$, and $\vec{b}_{3}$. The problem states that the projection of an unknown vector $\vec{a}$ onto each of these three vectors is equal. We recall the formula for the projection of vector $\vec{a}$ onto vector $\vec{b}$, which is given by the dot product of $\vec{a}$ and $\vec{b}$ divided by the magnitude of $\vec{b}$.

Step 2: Calculate magnitudes of the given vectors

To use the projection formula, we need the magnitudes of the vectors $\vec{b}_{1}$, $\vec{b}_{2}$, and $\vec{b}_{3}$. We calculate these magnitudes using the formula $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. We find that $|\vec{b}_{1}|$ is 3, $|\vec{b}_{2}|$ is 3, and $|\vec{b}_{3}|$ is 1.

Step 3: Set up equations based on equal projections

The problem states that the projections are equal. We set up the equality using the projection formula and the calculated magnitudes. Let $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$.

Step 4: Solve for the components of vector $\vec{a}$

We equate the first two projections and the second and third projections to form a system of linear equations. Solving these equations for $x, y, z$ in terms of a common ratio, we find that $x$, $y$, and $z$ are proportional to 7, 9, and 5 respectively. This means $\vec{a}$ can be written as $k(7\hat{i} + 9\hat{j} + 5\hat{k})$ for some scalar $k$.

Step 5: Find the unit vector along $\vec{a}$

Since $\vec{a}$ is a nonzero vector, $k \neq 0$. The unit vector along $\vec{a}$ is found by dividing $\vec{a}$ by its magnitude. This gives us two possible unit vectors, differing only by a sign, as the problem does not specify a direction for $\vec{a}$.

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