Key Concepts and Formulas

• Vector Cross Product: The cross product of two vectors $\overrightarrow{\mathrm{A}} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\overrightarrow{\mathrm{B}} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$ is given by: $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_y B_z - A_z B_y)\hat{i} - (A_x B_z - A_z B_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k}$
• Scalar Projection: The scalar projection of a vector $\overrightarrow{\mathrm{P}}$ onto a vector $\overrightarrow{\mathrm{Q}}$ is given by $\frac{\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}}{|\overrightarrow{\mathrm{Q}}|}$.
• Dot Product: The dot product of two vectors $\overrightarrow{\mathrm{A}} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\overrightarrow{\mathrm{B}} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$ is $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}} = A_x B_x + A_y B_y + A_z B_z$.
• Magnitude of a Vector: The magnitude of a vector $\overrightarrow{\mathrm{V}} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}$ is $|\overrightarrow{\mathrm{V}}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$.

Step-by-Step Solution

We are given $\overrightarrow{\mathrm{a}}=\alpha \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}-\alpha \hat{k}$, with $\alpha>0$. The projection of $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ on the vector $-\hat{i}+2 \hat{j}-2 \hat{k}$ is 30.

Step 1: Calculate the Cross Product $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$

• Why: The problem requires us to first find the vector $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ before we can project it.
• Using the determinant formula for the cross product: $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & -1 \\ 2 & 1 & -\alpha \end{vmatrix}$
• Expanding the determinant: $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = \hat{i}((1)(-\alpha) - (-1)(1)) - \hat{j}((\alpha)(-\alpha) - (-1)(2)) + \hat{k}((\alpha)(1) - (1)(2))$ $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = \hat{i}(-\alpha + 1) - \hat{j}(-\alpha^2 + 2) + \hat{k}(\alpha - 2)$ $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = (1-\alpha)\hat{i} + (\alpha^2-2)\hat{j} + (\alpha-2)\hat{k}$

Step 2: Identify the Vector for Projection and Calculate its Magnitude

• Why: To use the scalar projection formula, we need the vector onto which we are projecting and its magnitude.
• Let $\overrightarrow{\mathrm{c}} = -\hat{i}+2 \hat{j}-2 \hat{k}$.
• Calculate the magnitude of $\overrightarrow{\mathrm{c}}$: $|\overrightarrow{\mathrm{c}}| = \sqrt{(-1)^2 + (2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Step 3: Calculate the Dot Product of $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$

• Why: The scalar projection formula requires the dot product of the two vectors involved.
• Let $\overrightarrow{\mathrm{P}} = \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} = (1-\alpha)\hat{i} + (\alpha^2-2)\hat{j} + (\alpha-2)\hat{k}$ and $\overrightarrow{\mathrm{Q}} = \overrightarrow{\mathrm{c}} = -\hat{i}+2 \hat{j}-2 \hat{k}$.
• The dot product is: $\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}} = ((1-\alpha)(-1)) + ((\alpha^2-2)(2)) + ((\alpha-2)(-2))$ $\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}} = (-1 + \alpha) + (2\alpha^2 - 4) + (-2\alpha + 4)$ $\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}} = 2\alpha^2 - \alpha - 1$

Step 4: Apply the Scalar Projection Formula and Form an Equation

• Why: We are given that the scalar projection is 30. We can now set up an equation using the formula.
• The scalar projection of $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ on $\overrightarrow{\mathrm{c}}$ is $\frac{\overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}}{|\overrightarrow{\mathrm{Q}}|}$.
• We are given this value is 30: $\frac{2\alpha^2 - \alpha - 1}{3} = 30$

Step 5: Solve the Equation for $\alpha$}

• Why: This equation will allow us to find the value(s) of $\alpha$.
• Multiply both sides by 3: $2\alpha^2 - \alpha - 1 = 90$
• Rearrange into a quadratic equation: $2\alpha^2 - \alpha - 91 = 0$
• Factor the quadratic equation. We look for two numbers that multiply to $2 \times (-91) = -182$ and add to $-1$. These numbers are $-14$ and $13$. $2\alpha^2 - 14\alpha + 13\alpha - 91 = 0$ $2\alpha(\alpha - 7) + 13(\alpha - 7) = 0$ $(2\alpha + 13)(\alpha - 7) = 0$
• The possible values for $\alpha$ are: $2\alpha + 13 = 0 \implies \alpha = -\frac{13}{2}$ $\alpha - 7 = 0 \implies \alpha = 7$
• Apply the condition: The problem states that $\alpha > 0$. Therefore, we discard $\alpha = -\frac{13}{2}$.
• The valid solution is $\alpha = 7$.

Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with the signs when expanding the determinant for the cross product, especially the negative sign associated with the $\hat{j}$ component.
• Forgetting the Magnitude: The scalar projection formula involves dividing by the magnitude of the projection vector. Ensure this magnitude is calculated correctly.
• Ignoring Conditions: Always check if there are any conditions on the variables (like $\alpha>0$) and use them to select the appropriate solution from the derived possibilities.

Summary

The problem required calculating the cross product of two given vectors, determining the magnitude of a third vector, and then finding the scalar projection of the cross product onto the third vector. By equating this projection to the given value of 30, we formed a quadratic equation in $\alpha$. Solving this equation and applying the condition $\alpha>0$ yielded the unique value of $\alpha$.

The final answer is $\boxed{7}$.