Question

Obtain the projection of the vector \( \vec{a} = 2\hat{i} + 3\hat{j} + 5\hat{k} \) on the vector \( \vec{b} = \hat{i} + 3\hat{j} + \hat{k} \).

Hint:

Remember the distinction: The "scalar projection" is a length (a scalar), calculated as \( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \). The "vector projection" is a vector, calculated as \( \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b} \). Pay close attention to what the question asks for.

Answer 1

Step 1: Understanding the Concept:
The projection of a vector \( \vec{a} \) onto another vector \( \vec{b} \) is the scalar length of the component of \( \vec{a} \) that lies in the direction of \( \vec{b} \).
Step 2: Key Formula or Approach:
The formula for the scalar projection of vector \( \vec{a} \) on vector \( \vec{b} \) is given by: \[ \text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \] We need to compute the dot product \( \vec{a} \cdot \vec{b} \) and the magnitude of \( \vec{b} \).
Step 3: Detailed Explanation:
Given vectors are: \[ \vec{a} = 2\hat{i} + 3\hat{j} + 5\hat{k} \] \[ \vec{b} = \hat{i} + 3\hat{j} + \hat{k} \] First, calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (2)(1) + (3)(3) + (5)(1) \] \[ \vec{a} \cdot \vec{b} = 2 + 9 + 5 = 16 \] Next, calculate the magnitude of \( \vec{b} \): \[ |\vec{b}| = \sqrt{(1)^2 + (3)^2 + (1)^2} \] \[ |\vec{b}| = \sqrt{1 + 9 + 1} = \sqrt{11} \] Now, substitute these values into the projection formula: \[ \text{Projection} = \frac{16}{\sqrt{11}} \] Step 4: Final Answer:
The projection of the vector \( \vec{a} \) on the vector \( \vec{b} \) is \( \frac{16}{\sqrt{11}} \).