To solve the problem of finding the projection of vector $\vec{p}$ on vector $\vec{q}$, let's break it down step-by-step.

We are given: $\vec{p} = 5\hat{i} - 3\hat{j} + 2\hat{k}$ $\vec{q} = 2\hat{i} + 4\hat{j} + 2\hat{k}$

Formula for the projection of $\vec{p}$ on $\vec{q}$:

$\text{Projection of } \vec{p} \text{ on } \vec{q} = \frac{\vec{p} \cdot \vec{q}}{\|\vec{q}\|^2} \vec{q}$

Step 1: Dot product $\vec{p} \cdot \vec{q}$

$\vec{p} \cdot \vec{q} = (5)(2) + (-3)(4) + (2)(2) = 10 - 12 + 4 = 2$

Step 2: Magnitude squared of $\vec{q}$ ($\|\vec{q}\|^2$)

$\|\vec{q}\|^2 = (2)^2 + (4)^2 + (2)^2 = 4 + 16 + 4 = 24$

Step 3: Projection calculation

$\text{Projection of } \vec{p} \text{ on } \vec{q} = \frac{2}{24} \vec{q} = \frac{1}{12} \vec{q}$

Step 4: Substitute $\vec{q}$

$\frac{1}{12} \vec{q} = \frac{1}{12}(2\hat{i} + 4\hat{j} + 2\hat{k}) = \frac{2}{12}\hat{i} + \frac{4}{12}\hat{j} + \frac{2}{12}\hat{k}$ $= \frac{1}{6}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{6}\hat{k}$

Final Answer:

$\text{Projection of } \vec{p} \text{ on } \vec{q} = \frac{1}{6}\hat{i} + \frac{1}{3}\hat{j} + \frac{1}{6}\hat{k}$

Would you like further clarification or assistance with this solution?

Related Questions:

1. What is the angle between vectors $\vec{p}$ and $\vec{q}$?
2. How do you calculate the unit vector of $\vec{q}$?
3. What is the orthogonal projection of $\vec{p}$ onto a plane normal to $\vec{q}$?
4. How can you generalize the projection formula for n-dimensional vectors?
5. What is the physical interpretation of projecting one vector onto another?

Tip:

Always verify calculations for the dot product and magnitude carefully to avoid minor errors in vector operations!