Key Concepts and Formulas

• Cross Product of Vectors: For $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, $\vec{A} \times \vec{B} = (A_y B_z - A_z B_y)\hat{i} - (A_x B_z - A_z B_x)\hat{j} + (A_x B_y - A_y B_x)\hat{k}$
• Vector Equality: Two vectors are equal if and only if their corresponding components are equal.
• Vector Arithmetic: Vector addition, subtraction, and scalar multiplication are performed component-wise.
• Dot Product: For $\vec{P} = P_x \hat{i} + P_y \hat{j} + P_z \hat{k}$ and $\vec{Q} = Q_x \hat{i} + Q_y \hat{j} + Q_z \hat{k}$, $\vec{P} \cdot \vec{Q} = P_x Q_x + P_y Q_y + P_z Q_z$.
• Magnitude of a Vector: For $\vec{Q} = Q_x \hat{i} + Q_y \hat{j} + Q_z \hat{k}$, $|\vec{Q}| = \sqrt{Q_x^2 + Q_y^2 + Q_z^2}$.
• Scalar Projection: The scalar projection of vector $\vec{P}$ onto vector $\vec{Q}$ is given by $\text{Proj}_{\vec{Q}} \vec{P} = \frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}$.

Step-by-Step Solution

Step 1: Determine the Unknown Components $\alpha$ and $\beta$ of Vector $\vec{a}$

We are given $\vec{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}+4 \hat{k}$, and their cross product $\vec{a} \times \vec{b}=-\hat{i}+9 \hat{j}+12 \hat{k}$. We will use the cross product formula to find $\alpha$ and $\beta$.

First, we compute the cross product of $\vec{a}$ and $\vec{b}$: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & \beta \\ 3 & -5 & 4 \end{vmatrix}$ Expanding the determinant: $\vec{a} \times \vec{b} = (1 \cdot 4 - \beta \cdot (-5))\hat{i} - (\alpha \cdot 4 - \beta \cdot 3)\hat{j} + (\alpha \cdot (-5) - 1 \cdot 3)\hat{k}$ $\vec{a} \times \vec{b} = (4 + 5\beta)\hat{i} - (4\alpha - 3\beta)\hat{j} + (-5\alpha - 3)\hat{k}$ $\vec{a} \times \vec{b} = (4 + 5\beta)\hat{i} + (3\beta - 4\alpha)\hat{j} + (-5\alpha - 3)\hat{k}$

Now, we equate the components of this result with the given cross product, $-\hat{i}+9 \hat{j}+12 \hat{k}$:

• Equating the $\hat{i}$ components: $4 + 5\beta = -1$ $5\beta = -5$ $\beta = -1$

• Equating the $\hat{k}$ components: $-5\alpha - 3 = 12$ $-5\alpha = 15$ $\alpha = -3$

To verify, we check the $\hat{j}$ components: $3\beta - 4\alpha = 3(-1) - 4(-3) = -3 + 12 = 9$ This matches the given $\hat{j}$ component. Thus, our values for $\alpha$ and $\beta$ are correct.

Now we can write the complete vector $\vec{a}$: $\vec{a} = -3\hat{i} + \hat{j} - \hat{k}$

Step 2: Construct the Vectors Required for Projection

We need to find the projection of $\vec{b}-2\vec{a}$ on $\vec{b}+\vec{a}$. Let's first calculate these two vectors.

We have $\vec{a} = -3\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$.

1. Calculate $\vec{b}-2\vec{a}$: First, find $2\vec{a}$: $2\vec{a} = 2(-3\hat{i} + \hat{j} - \hat{k}) = -6\hat{i} + 2\hat{j} - 2\hat{k}$ Now, subtract $2\vec{a}$ from $\vec{b}$: $\vec{b}-2\vec{a} = (3\hat{i} - 5\hat{j} + 4\hat{k}) - (-6\hat{i} + 2\hat{j} - 2\hat{k})$ $\vec{b}-2\vec{a} = (3 - (-6))\hat{i} + (-5 - 2)\hat{j} + (4 - (-2))\hat{k}$ $\vec{b}-2\vec{a} = 9\hat{i} - 7\hat{j} + 6\hat{k}$

2. Calculate $\vec{b}+\vec{a}$: Add $\vec{a}$ to $\vec{b}$: $\vec{b}+\vec{a} = (3\hat{i} - 5\hat{j} + 4\hat{k}) + (-3\hat{i} + \hat{j} - \hat{k})$ $\vec{b}+\vec{a} = (3 + (-3))\hat{i} + (-5 + 1)\hat{j} + (4 + (-1))\hat{k}$ $\vec{b}+\vec{a} = 0\hat{i} - 4\hat{j} + 3\hat{k}$ $\vec{b}+\vec{a} = -4\hat{j} + 3\hat{k}$

Step 3: Calculate the Scalar Projection

We need to find the scalar projection of $\vec{P} = \vec{b}-2\vec{a} = 9\hat{i} - 7\hat{j} + 6\hat{k}$ onto $\vec{Q} = \vec{b}+\vec{a} = -4\hat{j} + 3\hat{k}$. The formula for scalar projection is $\text{Proj}_{\vec{Q}} \vec{P} = \frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}$.

1. Calculate the dot product $\vec{P} \cdot \vec{Q}$: $\vec{P} \cdot \vec{Q} = (9\hat{i} - 7\hat{j} + 6\hat{k}) \cdot (0\hat{i} - 4\hat{j} + 3\hat{k})$ $\vec{P} \cdot \vec{Q} = (9)(0) + (-7)(-4) + (6)(3)$ $\vec{P} \cdot \vec{Q} = 0 + 28 + 18$ $\vec{P} \cdot \vec{Q} = 46$

2. Calculate the magnitude of $\vec{Q}$: $|\vec{Q}| = |-4\hat{j} + 3\hat{k}|$ $|\vec{Q}| = \sqrt{(0)^2 + (-4)^2 + (3)^2}$ $|\vec{Q}| = \sqrt{0 + 16 + 9}$ $|\vec{Q}| = \sqrt{25}$ $|\vec{Q}| = 5$

3. Compute the scalar projection: $\text{Proj}_{\vec{b}+\vec{a}} (\vec{b}-2\vec{a}) = \frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|} = \frac{46}{5}$

Common Mistakes & Tips

• Sign Errors in Cross Product: Be meticulous when calculating the determinant for the cross product, as sign errors are common and can propagate through the entire solution.
• Component-wise Operations: Ensure that vector addition, subtraction, and scalar multiplication are performed correctly on each component.
• Projection Formula: Remember that the scalar projection of $\vec{P}$ onto $\vec{Q}$ is $\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}$, and not the other way around. The denominator is the magnitude of the vector onto which the projection is made.

Summary

The problem required us to first find the unknown components of vector $\vec{a}$ by utilizing the given cross product. Once $\vec{a}$ was fully determined, we proceeded to construct the vectors $\vec{b}-2\vec{a}$ and $\vec{b}+\vec{a}$ through vector arithmetic. Finally, we calculated the scalar projection of the former onto the latter by computing their dot product and the magnitude of the latter, and then dividing the dot product by the magnitude.

The final answer is $\boxed{\frac{46}{5}}$.