Key Concepts and Formulas

1. Cross Product of Vectors: The cross product of two vectors $\vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ and $\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}$ is given by: $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = (u_2 v_3 - u_3 v_2)\hat{i} - (u_1 v_3 - u_3 v_1)\hat{j} + (u_1 v_2 - u_2 v_1)\hat{k}$ The resultant vector is perpendicular to both $\vec{u}$ and $\vec{v}$.

2. Dot Product of Vectors: The dot product of $\vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ and $\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}$ is: $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$ This operation results in a scalar.

3. Magnitude of a Vector: The magnitude of a vector $\vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ is: $|\vec{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}$

4. Scalar Projection: The scalar projection of vector $\vec{x}$ onto vector $\vec{y}$ is the component of $\vec{x}$ in the direction of $\vec{y}$, given by: $\text{Proj}_{\vec{y}} \vec{x} = \frac{\vec{x} \cdot \vec{y}}{|\vec{y}|}$

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Step-by-Step Solution

We are given $\vec{a}=3 \hat{i}-\hat{j}+2 \hat{k}$. We need to find the projection of $\vec{c}-2 \hat{j}$ on $\vec{a}$. This requires us to first compute $\vec{b}$ and then $\vec{c}$.

Step 1: Calculate vector $\vec{b}$ We are given $\vec{b} = \vec{a} \times (\hat{i}-2 \hat{k})$. We substitute the components of $\vec{a}$ and the second vector. $\vec{a} = 3\hat{i} - 1\hat{j} + 2\hat{k}$ $\hat{i}-2 \hat{k} = 1\hat{i} + 0\hat{j} - 2\hat{k}$

Using the determinant formula for the cross product: $\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix}$ Expanding the determinant: $\vec{b} = \hat{i}((-1)(-2) - (2)(0)) - \hat{j}((3)(-2) - (2)(1)) + \hat{k}((3)(0) - (-1)(1))$ $\vec{b} = \hat{i}(2 - 0) - \hat{j}(-6 - 2) + \hat{k}(0 + 1)$ $\vec{b} = 2\hat{i} - (-8)\hat{j} + 1\hat{k}$ $\vec{b} = 2\hat{i} + 8\hat{j} + \hat{k}$ Reasoning: This step applies the definition of the cross product to find the vector $\vec{b}$ from the given $\vec{a}$ and another vector.

Step 2: Calculate vector $\vec{c}$ We are given $\vec{c} = \vec{b} \times \hat{k}$. We use the components of $\vec{b}$ calculated in the previous step and the components of $\hat{k}$. $\vec{b} = 2\hat{i} + 8\hat{j} + 1\hat{k}$ $\hat{k} = 0\hat{i} + 0\hat{j} + 1\hat{k}$

Using the determinant formula for the cross product: $\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{vmatrix}$ Expanding the determinant: $\vec{c} = \hat{i}((8)(1) - (1)(0)) - \hat{j}((2)(1) - (1)(0)) + \hat{k}((2)(0) - (8)(0))$ $\vec{c} = \hat{i}(8 - 0) - \hat{j}(2 - 0) + \hat{k}(0 - 0)$ $\vec{c} = 8\hat{i} - 2\hat{j}$ Reasoning: This step calculates $\vec{c}$ by performing another cross product, this time between $\vec{b}$ and the unit vector $\hat{k}$.

Step 3: Calculate the vector $(\vec{c} - 2\hat{j})$ We need to find the vector that we will project onto $\vec{a}$. This vector is $\vec{c} - 2\hat{j}$. $\vec{c} = 8\hat{i} - 2\hat{j}$ $\vec{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j}$ $\vec{c} - 2\hat{j} = 8\hat{i} - 4\hat{j}$ For the dot product calculation, we can explicitly write the $\hat{k}$ component as 0: $8\hat{i} - 4\hat{j} + 0\hat{k}$. Reasoning: This step combines the vector $\vec{c}$ with the given vector $-2\hat{j}$ through vector subtraction.

Step 4: Calculate the magnitude of $\vec{a}$ The scalar projection formula requires the magnitude of the vector onto which we are projecting, which is $\vec{a}$. $\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}$ $|\vec{a}| = \sqrt{(3)^2 + (-1)^2 + (2)^2}$ $|\vec{a}| = \sqrt{9 + 1 + 4}$ $|\vec{a}| = \sqrt{14}$ Reasoning: This step finds the length of the vector $\vec{a}$, which is a necessary component for the scalar projection calculation.

Step 5: Calculate the dot product of $(\vec{c} - 2\hat{j})$ and $\vec{a}$ To find the scalar projection, we need the dot product of the vector to be projected $(\vec{c} - 2\hat{j})$ and the vector it's projected onto $\vec{a}$. $(\vec{c} - 2\hat{j}) = 8\hat{i} - 4\hat{j} + 0\hat{k}$ $\vec{a} = 3\hat{i} - 1\hat{j} + 2\hat{k}$ $(\vec{c} - 2\hat{j}) \cdot \vec{a} = (8)(3) + (-4)(-1) + (0)(2)$ $(\vec{c} - 2\hat{j}) \cdot \vec{a} = 24 + 4 + 0$ $(\vec{c} - 2\hat{j}) \cdot \vec{a} = 28$ Reasoning: This step computes the dot product, which measures the extent to which the two vectors align.

Step 6: Calculate the scalar projection of $(\vec{c} - 2\hat{j})$ on $\vec{a}$ Now we use the scalar projection formula with the results from Step 5 and Step 4. $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = \frac{(\vec{c} - 2\hat{j}) \cdot \vec{a}}{|\vec{a}|}$ $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = \frac{28}{\sqrt{14}}$ To simplify, we can multiply the numerator and denominator by $\sqrt{14}$: $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = \frac{28 \sqrt{14}}{14}$ $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = 2\sqrt{14}$ Correction: Let's recheck the calculation. $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix} = \hat{i}(2-0) - \hat{j}(-6-2) + \hat{k}(0-(-1)) = 2\hat{i} + 8\hat{j} + \hat{k}$ $\vec{c} = \vec{b} \times \hat{k} = (2\hat{i} + 8\hat{j} + \hat{k}) \times \hat{k} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{vmatrix} = \hat{i}(8-0) - \hat{j}(2-0) + \hat{k}(0-0) = 8\hat{i} - 2\hat{j}$ $\vec{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j} = 8\hat{i} - 4\hat{j}$ We need the projection of $\vec{c} - 2\hat{j}$ on $\vec{a}$. Projection $= \frac{(\vec{c} - 2\hat{j}) \cdot \vec{a}}{|\vec{a}|}$ $(\vec{c} - 2\hat{j}) \cdot \vec{a} = (8\hat{i} - 4\hat{j}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) = (8)(3) + (-4)(-1) + (0)(2) = 24 + 4 + 0 = 28$. $|\vec{a}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9+1+4} = \sqrt{14}$. Projection $= \frac{28}{\sqrt{14}} = \frac{28\sqrt{14}}{14} = 2\sqrt{14}$.

There seems to be a discrepancy with the provided correct answer. Let's re-evaluate the problem statement and calculations.

Let's assume the question or options might have a typo, and re-examine the calculations for any potential errors. The steps followed are standard for vector operations.

Let's double check the cross product of $\vec{a}$ and $(\hat{i} - 2\hat{k})$. $\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}$ $\vec{v} = \hat{i} - 2\hat{k}$ $\vec{a} \times \vec{v} = ((-1)(-2) - (2)(0))\hat{i} - ((3)(-2) - (2)(1))\hat{j} + ((3)(0) - (-1)(1))\hat{k}$ $= (2-0)\hat{i} - (-6-2)\hat{j} + (0-(-1))\hat{k}$ $= 2\hat{i} + 8\hat{j} + \hat{k}$. This matches.

Now, $\vec{c} = \vec{b} \times \hat{k}$. $\vec{b} = 2\hat{i} + 8\hat{j} + \hat{k}$ $\vec{c} = (2\hat{i} + 8\hat{j} + \hat{k}) \times \hat{k}$ $= (8(1) - 1(0))\hat{i} - (2(1) - 1(0))\hat{j} + (2(0) - 8(0))\hat{k}$ $= 8\hat{i} - 2\hat{j}$. This matches.

Now, $\vec{c} - 2\hat{j} = 8\hat{i} - 2\hat{j} - 2\hat{j} = 8\hat{i} - 4\hat{j}$. This matches.

Projection of $8\hat{i} - 4\hat{j}$ on $\vec{a} = 3\hat{i} - \hat{j} + 2\hat{k}$. Dot product: $(8\hat{i} - 4\hat{j}) \cdot (3\hat{i} - \hat{j} + 2\hat{k}) = 8(3) + (-4)(-1) + 0(2) = 24 + 4 + 0 = 28$. This matches. Magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9+1+4} = \sqrt{14}$. This matches. Projection: $\frac{28}{\sqrt{14}} = \frac{28\sqrt{14}}{14} = 2\sqrt{14}$.

Let's check if there's a property of cross products that could simplify this, but the direct calculation is usually the most reliable.

Let's consider the possibility of a typo in the question or options. If the answer is indeed $2\sqrt{7}$, then either the dot product should be $14$ or the magnitude of $\vec{a}$ should be $\sqrt{28} = 2\sqrt{7}$. Neither seems to be the case.

Let's re-examine the problem and options. Options: (A) $2 \sqrt{7}$, (B) $3 \sqrt{7}$, (C) $\sqrt{14}$, (D) $2 \sqrt{14}$.

Our calculated value is $2\sqrt{14}$. This matches option (D). However, the provided correct answer is (A) $2\sqrt{7}$. This indicates a strong possibility of an error in the provided "Correct Answer".

Assuming our calculations are correct, $2\sqrt{14}$ is the result. Let's proceed with the assumption that our calculation is correct and the provided answer might be wrong.

Step 6 (Revised): Calculate the scalar projection of $(\vec{c} - 2\hat{j})$ on $\vec{a}$ Using the scalar projection formula with the results from Step 5 and Step 4. $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = \frac{(\vec{c} - 2\hat{j}) \cdot \vec{a}}{|\vec{a}|}$ $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = \frac{28}{\sqrt{14}}$ To simplify, we rationalize the denominator: $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = \frac{28 \sqrt{14}}{14}$ $\text{Proj}_{\vec{a}} (\vec{c} - 2\hat{j}) = 2\sqrt{14}$ Reasoning: This step completes the calculation of the scalar projection by dividing the dot product by the magnitude of the vector $\vec{a}$.

Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with the signs when expanding the determinant for the cross product. The middle term for $\hat{j}$ is always subtracted.
• Order of Vectors in Cross Product: The cross product is not commutative ($\vec{u} \times \vec{v} \neq \vec{v} \times \vec{u}$). Ensure you are using the vectors in the correct order as specified.
• Scalar vs. Vector Projection: The question asks for the "projection of $\vec{c}-2 \hat{j}$ on $\vec{a}$". This implies scalar projection (a length), as opposed to vector projection (a vector). The formula used is for scalar projection.
• Rationalization: Always rationalize the denominator for the final answer if it involves a square root.

Summary

The problem requires a series of vector operations: cross products to find vectors $\vec{b}$ and $\vec{c}$, followed by vector subtraction to find the vector to be projected. Finally, the scalar projection formula is applied using the dot product and magnitude of the involved vectors. Our calculations yielded $2\sqrt{14}$.

The final answer is $\boxed{2\sqrt{14}}$. This corresponds to option (D). However, given the provided correct answer is (A) $2\sqrt{7}$, there might be an error in the problem statement or the provided answer key. Based on standard vector algebra, the calculation leads to $2\sqrt{14}$.