1. Key Concepts and Formulas

• Scalar Projection of Vector $\vec{a}$ on Vector $\vec{b}$: The scalar projection of vector $\vec{a}$ onto vector $\vec{b}$ is given by the formula: $\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ This value represents the signed length of the projection of $\vec{a}$ onto the line containing $\vec{b}$.

• Dot Product of Vectors: For two vectors $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$, their dot product is: $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$

• Magnitude of a Vector: For a vector $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$, its magnitude is: $|\vec{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2}$

• Direction of Projection:

  • If $\vec{a} \cdot \vec{b} > 0$, the projection vector is in the same direction as $\vec{b}$.
  • If $\vec{a} \cdot \vec{b} < 0$, the projection vector is in the opposite direction to $\vec{b}$.
  • If $\vec{a} \cdot \vec{b} = 0$, the vectors are orthogonal, and the projection is the zero vector.

2. Step-by-Step Solution

We are given the vectors $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}+5 \hat{k}$. We need to find the scalar projection of $\vec{a}$ on $\vec{b}$ and determine the direction of the projection vector.

Step 1: Calculate the dot product $\vec{a} \cdot \vec{b}$ The dot product measures the extent to which two vectors point in the same direction. $\vec{a} \cdot \vec{b} = (5 \hat{i} - \hat{j} - 3 \hat{k}) \cdot (\hat{i} + 3 \hat{j} + 5 \hat{k})$ Multiply the corresponding components: $\vec{a} \cdot \vec{b} = (5)(1) + (-1)(3) + (-3)(5)$ $\vec{a} \cdot \vec{b} = 5 - 3 - 15$ $\vec{a} \cdot \vec{b} = -13$ Since the dot product is negative, it indicates that the angle between $\vec{a}$ and $\vec{b}$ is obtuse, and therefore the projection vector will be in the opposite direction to $\vec{b}$.

Step 2: Calculate the magnitude of vector $\vec{b}$, $|\vec{b}|$ The magnitude of $\vec{b}$ is its length, which will be the denominator in our projection formula. $|\vec{b}| = \sqrt{(1)^2 + (3)^2 + (5)^2}$ $|\vec{b}| = \sqrt{1 + 9 + 25}$ $|\vec{b}| = \sqrt{35}$

Step 3: Calculate the scalar projection of $\vec{a}$ on $\vec{b}$ Using the formula for scalar projection: $\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ Substitute the values calculated in Step 1 and Step 2: $\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{-13}{\sqrt{35}}$

Step 4: Determine the direction of the projection vector From Step 1, we found that $\vec{a} \cdot \vec{b} = -13$. Since the dot product is negative, the projection of $\vec{a}$ onto $\vec{b}$ points in the direction opposite to $\vec{b}$.

Step 5: Compare with the given options Our calculations show that the projection of $\vec{a}$ on $\vec{b}$ is $\frac{-13}{\sqrt{35}}$, and its direction is opposite to the direction of $\vec{b}$. This matches option (A).

• Option (A): Projection of $\vec{a}$ on $\vec{b}$ is $\frac{-13}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$. (Matches our result)
• Option (B): Projection of $\vec{a}$ on $\vec{b}$ is $\frac{13}{\sqrt{35}}$ (Incorrect value) and the direction of the projection vector is opposite to the direction of $\vec{b}$.
• Option (C): Projection of $\vec{a}$ on $\vec{b}$ is $\frac{13}{\sqrt{35}}$ (Incorrect value) and the direction of the projection vector is same as of $\vec{b}$.
• Option (D): Projection of $\vec{a}$ on $\vec{b}$ is $\frac{-13}{\sqrt{35}}$ and the direction of the projection vector is same as of $\vec{b}$. (Incorrect direction)

3. Common Mistakes & Tips

• Sign Error: Do not ignore the sign of the dot product. A negative dot product is crucial for determining the direction of the projection. It indicates an obtuse angle between the vectors.
• Magnitude Calculation: Ensure that the magnitude of the vector you are projecting onto is calculated correctly using the Pythagorean theorem for vectors.
• Scalar vs. Vector Projection: Remember that this question asks for the scalar projection (a signed number), not the vector projection (a vector).

4. Summary

To find the scalar projection of vector $\vec{a}$ on vector $\vec{b}$, we first computed their dot product, $\vec{a} \cdot \vec{b}$, and the magnitude of $\vec{b}$, $|\vec{b}|$. The scalar projection is then $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. The sign of the dot product determines the direction of the projection vector. A negative dot product means the projection vector is in the opposite direction to $\vec{b}$. In this case, $\vec{a} \cdot \vec{b} = -13$ and $|\vec{b}| = \sqrt{35}$, leading to a scalar projection of $\frac{-13}{\sqrt{35}}$. Since the dot product is negative, the direction of the projection vector is opposite to that of $\vec{b}$.

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