Key Concepts and Formulas

• Projection of a vector: The scalar projection of vector $\overrightarrow a$ onto vector $\overrightarrow c$ is given by $\text{Proj}_{\overrightarrow c} \overrightarrow a = \frac{\overrightarrow a \cdot \overrightarrow c}{|\overrightarrow c|}$.
• Dot Product: For vectors $\overrightarrow u = u_x \widehat i + u_y \widehat j + u_z \widehat k$ and $\overrightarrow v = v_x \widehat i + v_y \widehat j + v_z \widehat k$, their dot product is $\overrightarrow u \cdot \overrightarrow v = u_x v_x + u_y v_y + u_z v_z$.
• Magnitude of a vector: The magnitude of a vector $\overrightarrow v = v_x \widehat i + v_y \widehat j + v_z \widehat k$ is $|\overrightarrow v| = \sqrt{v_x^2 + v_y^2 + v_z^2}$.
• Cross Product: The cross product of two vectors $\overrightarrow b = b_x \widehat i + b_y \widehat j + b_z \widehat k$ and $\overrightarrow c = c_x \widehat i + c_y \widehat j + c_z \widehat k$ can be calculated using the determinant: $\overrightarrow b \times \overrightarrow c = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ {b_x} & {b_y} & {b_z} \\ {c_x} & {c_y} & {c_z} \\ \end{matrix} } \right|$

Step-by-Step Solution

Step 1: Determine the value of $\alpha$ using the given vector projection. We are given that the projection of $\overrightarrow a$ on $\overrightarrow c$ is $\frac{10}{3}$. The formula for scalar projection is $\text{Proj}_{\overrightarrow c} \overrightarrow a = \frac{\overrightarrow a \cdot \overrightarrow c}{|\overrightarrow c|}$. First, we calculate the dot product $\overrightarrow a \cdot \overrightarrow c$: $\overrightarrow a \cdot \overrightarrow c = (\alpha \widehat i + 3\widehat j - \widehat k) \cdot (\widehat i + 2\widehat j - 2\widehat k)$ $\overrightarrow a \cdot \overrightarrow c = (\alpha)(1) + (3)(2) + (-1)(-2)$ $\overrightarrow a \cdot \overrightarrow c = \alpha + 6 + 2$ $\overrightarrow a \cdot \overrightarrow c = \alpha + 8$ Next, we calculate the magnitude of $\overrightarrow c$: $|\overrightarrow c| = |\widehat i + 2\widehat j - 2\widehat k|$ $|\overrightarrow c| = \sqrt{1^2 + 2^2 + (-2)^2}$ $|\overrightarrow c| = \sqrt{1 + 4 + 4}$ $|\overrightarrow c| = \sqrt{9}$ $|\overrightarrow c| = 3$ Now, we use the given projection value: $\frac{\overrightarrow a \cdot \overrightarrow c}{|\overrightarrow c|} = \frac{10}{3}$ $\frac{\alpha + 8}{3} = \frac{10}{3}$ Multiplying both sides by 3, we get: $\alpha + 8 = 10$ $\alpha = 10 - 8$ $\alpha = 2$

Step 2: Determine the value of $\beta$ using the given cross product. We are given that $\overrightarrow b \times \overrightarrow c = - 6\widehat i + 10\widehat j + 7\widehat k$. We will compute the cross product of $\overrightarrow b$ and $\overrightarrow c$ using the determinant formula: $\overrightarrow b \times \overrightarrow c = \left| {\begin{matrix} {\widehat i} & {\widehat j} & {\widehat k} \\ 3 & { - \beta } & 4 \\ 1 & 2 & { - 2} \\ \end{matrix} } \right|$ Expanding the determinant: $\overrightarrow b \times \overrightarrow c = \widehat i ((- \beta)(-2) - (4)(2)) - \widehat j ((3)(-2) - (4)(1)) + \widehat k ((3)(2) - (- \beta)(1))$ $\overrightarrow b \times \overrightarrow c = \widehat i (2\beta - 8) - \widehat j (-6 - 4) + \widehat k (6 + \beta)$ $\overrightarrow b \times \overrightarrow c = (2\beta - 8)\widehat i - (-10)\widehat j + (6 + \beta)\widehat k$ $\overrightarrow b \times \overrightarrow c = (2\beta - 8)\widehat i + 10\widehat j + (6 + \beta)\widehat k$ Now, we equate this result with the given cross product: $(2\beta - 8)\widehat i + 10\widehat j + (6 + \beta)\widehat k = - 6\widehat i + 10\widehat j + 7\widehat k$ By comparing the corresponding components, we can form equations: For the $\widehat i$ component: $2\beta - 8 = -6$ $2\beta = -6 + 8$ $2\beta = 2$ $\beta = 1$ For the $\widehat j$ component: $10 = 10$ This confirms consistency. For the $\widehat k$ component: $6 + \beta = 7$ $\beta = 7 - 6$ $\beta = 1$ Both components give $\beta = 1$.

Step 3: Calculate the value of $\alpha + \beta$. We have found $\alpha = 2$ and $\beta = 1$. Therefore, their sum is: $\alpha + \beta = 2 + 1$ $\alpha + \beta = 3$

Common Mistakes & Tips

• Projection Denominator: Always use the magnitude of the vector onto which you are projecting (i.e., $|\overrightarrow c|$ in this case), not the magnitude of the vector being projected ($\overrightarrow a$).
• Cross Product Signs: Be extremely careful with the signs when expanding the determinant for the cross product, especially the negative sign associated with the $\widehat j$ component.
• Component Equivalence: For two vectors to be equal, all their corresponding components ($\widehat i$, $\widehat j$, $\widehat k$) must be equal. This provides a way to set up equations to solve for unknown scalar values.

Summary

The problem required us to find two unknown scalar components, $\alpha$ and $\beta$, of given vectors. We first utilized the concept of vector projection to find $\alpha$ by calculating the dot product and magnitude of the relevant vectors and solving the resulting equation. Subsequently, we employed the cross product formula, expressed as a determinant, to find $\beta$ by equating the calculated cross product with the given vector. Finally, we summed the determined values of $\alpha$ and $\beta$ to arrive at the answer.

The final answer is \boxed{3}.