Key Concepts and Formulas

1. Dot Product for Perpendicular Vectors: Two non-zero vectors $\overrightarrow u$ and $\overrightarrow v$ are perpendicular if and only if their dot product is zero: $\overrightarrow u \cdot \overrightarrow v = 0$.
2. Cross Product: For vectors $\overrightarrow u = u_1\widehat i + u_2\widehat j + u_3\widehat k$ and $\overrightarrow v = v_1\widehat i + v_2\widehat j + v_3\widehat k$, their cross product is given by: $\overrightarrow u \times \overrightarrow v = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2v_3 - u_3v_2)\widehat i - (u_1v_3 - u_3v_1)\widehat j + (u_1v_2 - u_2v_1)\widehat k$
3. Equality of Vectors: Two vectors are equal if their corresponding components are equal.
4. Projection of a Vector: The scalar projection of vector $\overrightarrow a$ onto vector $\overrightarrow b$ is given by $\text{Proj}_{\overrightarrow b} \overrightarrow a = \frac{\overrightarrow a \cdot \overrightarrow b}{|\overrightarrow b|}$.

Step-by-Step Solution

Step 1: Represent the unknown vector $\overrightarrow a$

Let the unknown vector $\overrightarrow a$ be represented as $\overrightarrow a = a_1\widehat i + a_2\widehat j + a_3\widehat k$. We use this general form to find its specific components using the given conditions.

Step 2: Use the perpendicularity condition

We are given that $\overrightarrow a$ is perpendicular to the vector $3\widehat i + \frac{1}{2}\widehat j + 2\widehat k$. Using the dot product property for perpendicular vectors: $\overrightarrow a \cdot \left( 3\widehat i + \frac{1}{2}\widehat j + 2\widehat k \right) = 0$ $(a_1\widehat i + a_2\widehat j + a_3\widehat k) \cdot \left( 3\widehat i + \frac{1}{2}\widehat j + 2\widehat k \right) = 0$ $3a_1 + \frac{1}{2}a_2 + 2a_3 = 0$ Multiplying by 2 to clear the fraction, we get our first equation: $6a_1 + a_2 + 4a_3 = 0 \quad \text{(Equation 1)}$ This equation establishes a linear relationship between the components of $\overrightarrow a$ due to the orthogonality condition.

Step 3: Use the cross product condition

We are given $\overrightarrow a \times \left( 2\widehat i + \widehat k \right) = 2\widehat i - 13\widehat j - 4\widehat k$. Let's compute the cross product on the left side: $\overrightarrow a \times \left( 2\widehat i + \widehat k \right) = (a_1\widehat i + a_2\widehat j + a_3\widehat k) \times (2\widehat i + 0\widehat j + 1\widehat k)$ $= \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ a_1 & a_2 & a_3 \\ 2 & 0 & 1 \end{vmatrix}$ $= \widehat i(a_2 \cdot 1 - a_3 \cdot 0) - \widehat j(a_1 \cdot 1 - a_3 \cdot 2) + \widehat k(a_1 \cdot 0 - a_2 \cdot 2)$ $= a_2\widehat i - (a_1 - 2a_3)\widehat j - 2a_2\widehat k$ Now, we equate this to the given vector: $a_2\widehat i - (a_1 - 2a_3)\widehat j - 2a_2\widehat k = 2\widehat i - 13\widehat j - 4\widehat k$ By comparing the corresponding components, we get the following equations: From the $\widehat i$ component: $a_2 = 2 \quad \text{(Equation 2)}$ From the $\widehat j$ component: $-(a_1 - 2a_3) = -13 \implies a_1 - 2a_3 = 13 \quad \text{(Equation 3)}$ From the $\widehat k$ component: $-2a_2 = -4 \implies a_2 = 2$ Equation 2 and the $\widehat k$ component equation confirm that $a_2 = 2$.

Step 4: Solve the system of linear equations

We have the following system of equations:

1. $6a_1 + a_2 + 4a_3 = 0$
2. $a_2 = 2$
3. $a_1 - 2a_3 = 13$

Substitute $a_2 = 2$ from Equation 2 into Equation 1: $6a_1 + 2 + 4a_3 = 0$ $6a_1 + 4a_3 = -2$ Dividing by 2, we get: $3a_1 + 2a_3 = -1 \quad \text{(Equation 4)}$ Now we solve the system formed by Equation 3 and Equation 4: 3. $a_1 - 2a_3 = 13$ 4. $3a_1 + 2a_3 = -1$ Adding Equation 3 and Equation 4 to eliminate $a_3$: $(a_1 - 2a_3) + (3a_1 + 2a_3) = 13 + (-1)$ $4a_1 = 12$ $a_1 = 3$ Substitute $a_1 = 3$ into Equation 3: $3 - 2a_3 = 13$ $-2a_3 = 10$ $a_3 = -5$ Thus, the components of $\overrightarrow a$ are $a_1 = 3$, $a_2 = 2$, and $a_3 = -5$. So, $\overrightarrow a = 3\widehat i + 2\widehat j - 5\widehat k$.

Step 5: Calculate the projection of $\overrightarrow a$ on $2\widehat i + 2\widehat j + \widehat k$

Let $\overrightarrow b = 2\widehat i + 2\widehat j + \widehat k$. We need to find the projection of $\overrightarrow a$ onto $\overrightarrow b$. The projection is given by the formula: $\text{Proj}_{\overrightarrow b} \overrightarrow a = \frac{\overrightarrow a \cdot \overrightarrow b}{|\overrightarrow b|}$.

First, calculate the dot product $\overrightarrow a \cdot \overrightarrow b$: $\overrightarrow a \cdot \overrightarrow b = (3\widehat i + 2\widehat j - 5\widehat k) \cdot (2\widehat i + 2\widehat j + \widehat k)$ $= (3)(2) + (2)(2) + (-5)(1)$ $= 6 + 4 - 5$ $= 5$

Next, calculate the magnitude of $\overrightarrow b$: $|\overrightarrow b| = \sqrt{(2)^2 + (2)^2 + (1)^2}$ $= \sqrt{4 + 4 + 1}$ $= \sqrt{9}$ $= 3$

Finally, calculate the projection: $\text{Proj}_{\overrightarrow b} \overrightarrow a = \frac{\overrightarrow a \cdot \overrightarrow b}{|\overrightarrow b|} = \frac{5}{3}$

Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with the signs when calculating the determinant for the cross product, especially for the $\widehat j$ component.
• Algebraic Errors: Solving the system of linear equations requires careful algebraic manipulation. Double-check your substitutions and eliminations.
• Magnitude Calculation: Ensure you correctly apply the Pythagorean theorem for the magnitude of a vector.

Summary

The problem required us to find an unknown vector $\overrightarrow a$ using two conditions: one involving perpendicularity (dot product) and another involving a cross product. By representing $\overrightarrow a$ in terms of its components and setting up a system of linear equations from the given conditions, we solved for its components. Once $\overrightarrow a$ was determined, we calculated its scalar projection onto another given vector using the dot product and the magnitude of the projection vector.

The projection of vector $\overrightarrow a$ on the vector $2\widehat i + 2\widehat j + \widehat k$ is $\frac{5}{3}$.

The final answer is $\boxed{\frac{5}{3}}$ which corresponds to option (C).