Key Concepts and Formulas

• Scalar Projection: The scalar projection of vector $\vec{a}$ onto a non-zero vector $\vec{b}$ is given by $\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. This represents the signed length of the component of $\vec{a}$ along the direction of $\vec{b}$.
• Dot Product: For vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, the dot product is $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$.
• Magnitude of a Vector: For a vector $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$, the magnitude is $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$.
• Unit Vector: A unit vector in the direction of a non-zero vector $\vec{a}$ is given by $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.

Step-by-Step Solution

Step 1: Define the unknown vector and given vectors. Let the unknown non-zero vector be $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$. The three given vectors are: $\vec{v_1} = 2\hat{i} - \hat{j} + 2\hat{k}$ $\vec{v_2} = \hat{i} + 2\hat{j} - 2\hat{k}$ $\vec{v_3} = \hat{k}$

Step 2: Calculate the magnitudes of the given vectors. We need the magnitudes of $\vec{v_1}$, $\vec{v_2}$, and $\vec{v_3}$ for the scalar projection formula. $|\vec{v_1}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ $|\vec{v_2}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$ $|\vec{v_3}| = \sqrt{0^2 + 0^2 + 1^2} = \sqrt{1} = 1$

Step 3: Set up equations based on the condition of equal projections. The problem states that the scalar projections of $\vec{a}$ on $\vec{v_1}$, $\vec{v_2}$, and $\vec{v_3}$ are equal. Let this common projection value be $P$. Projection of $\vec{a}$ on $\vec{v_1}$: $\text{Proj}_{\vec{v_1}} \vec{a} = \frac{\vec{a} \cdot \vec{v_1}}{|\vec{v_1}|} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} + 2\hat{k})}{3} = \frac{2x - y + 2z}{3}$ Projection of $\vec{a}$ on $\vec{v_2}$: $\text{Proj}_{\vec{v_2}} \vec{a} = \frac{\vec{a} \cdot \vec{v_2}}{|\vec{v_2}|} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + 2\hat{j} - 2\hat{k})}{3} = \frac{x + 2y - 2z}{3}$ Projection of $\vec{a}$ on $\vec{v_3}$: $\text{Proj}_{\vec{v_3}} \vec{a} = \frac{\vec{a} \cdot \vec{v_3}}{|\vec{v_3}|} = \frac{(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{k})}{1} = \frac{z}{1} = z$ Equating these projections:

1. $\frac{2x - y + 2z}{3} = P$
2. $\frac{x + 2y - 2z}{3} = P$
3. $z = P$

Step 4: Solve the system of equations to find the relationship between x, y, and z. Substitute $P=z$ into equations (1) and (2): From equation (1): $2x - y + 2z = 3z$ $2x - y - z = 0 \quad (*)$

From equation (2): $x + 2y - 2z = 3z$ $x + 2y - 5z = 0 \quad (**)$

We now have a system of two linear equations with three variables. We can express $x$ and $y$ in terms of $z$. From equation $(*)$, isolate $y$: $y = 2x - z$

Substitute this expression for $y$ into equation $(**)$: $x + 2(2x - z) - 5z = 0$ $x + 4x - 2z - 5z = 0$ $5x - 7z = 0$ $x = \frac{7}{5}z$

Now substitute the value of $x$ back into the expression for $y$: $y = 2\left(\frac{7}{5}z\right) - z$ $y = \frac{14}{5}z - \frac{5}{5}z$ $y = \frac{9}{5}z$

So, the components of $\vec{a}$ are $x = \frac{7}{5}z$, $y = \frac{9}{5}z$, and $z=z$. This means $\vec{a} = \frac{7}{5}z\hat{i} + \frac{9}{5}z\hat{j} + z\hat{k}$.

Step 5: Express the vector $\vec{a}$ in a simpler form and find its direction. We can factor out $\frac{z}{5}$ from the expression for $\vec{a}$: $\vec{a} = \frac{z}{5} (7\hat{i} + 9\hat{j} + 5\hat{k})$ Since $\vec{a}$ is a non-zero vector, $z \neq 0$. Let $k = \frac{z}{5}$, where $k$ is a non-zero scalar. Then $\vec{a} = k(7\hat{i} + 9\hat{j} + 5\hat{k})$. This shows that $\vec{a}$ is parallel to the vector $7\hat{i} + 9\hat{j} + 5\hat{k}$.

Step 6: Calculate the unit vector along $\vec{a}$. A unit vector along $\vec{a}$ is found by dividing $\vec{a}$ by its magnitude. The unit vector will be in the same direction as $7\hat{i} + 9\hat{j} + 5\hat{k}$. Let $\vec{d} = 7\hat{i} + 9\hat{j} + 5\hat{k}$. First, calculate the magnitude of $\vec{d}$: $|\vec{d}| = \sqrt{7^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155}$

The unit vector along $\vec{a}$ is: $\hat{a} = \frac{\vec{d}}{|\vec{d}|} = \frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})$

Step 7: Compare with the given options. The calculated unit vector is $\frac{1}{\sqrt{155}}(7\hat{i} + 9\hat{j} + 5\hat{k})$. This matches option (A).

Common Mistakes & Tips

• Scalar vs. Vector Projection: Ensure you are using the scalar projection formula. The problem asks for "projections," which are scalar values.
• Algebraic Errors: Solving the system of linear equations requires careful attention to detail to avoid sign errors or miscalculations.
• Magnitude Calculation: Double-check the calculation of vector magnitudes to ensure accuracy.

Summary

The problem requires finding a unit vector along $\vec{a}$ given that its scalar projections on three different vectors are equal. We set up a system of equations by equating the scalar projection formula for each given vector. By solving this system, we determined the relationship between the components of $\vec{a}$, allowing us to express $\vec{a}$ as a scalar multiple of a specific vector. Finally, we calculated the unit vector in this direction.

The final answer is \boxed{\text{\frac{1}{\sqrt{155}}(-7 \hat{i}+9 \hat{j}+5 \hat{k})}}.