Question

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

Hint:

Distinguish between scalar projection and vector projection. Scalar projection is just the length, given by \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\). Vector projection is a vector, found by multiplying the scalar projection by the unit vector of \(\vec{b}\), which results in the formula \(\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}\). Read the question carefully to determine which is required.

Answer 1

Step 1: Understanding the Concept:
The projection of a vector \(\vec{a}\) onto a non-zero vector \(\vec{b}\) is the orthogonal projection of \(\vec{a}\) onto a straight line parallel to \(\vec{b}\). This results in a new vector that is parallel to \(\vec{b}\).
Step 2: Key Formula or Approach:
The vector projection of \(\vec{a}\) onto \(\vec{b}\) is given by the formula:
\[ \text{proj}_{\vec{b}} \vec{a} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b} \] This formula requires us to compute the dot product \(\vec{a} \cdot \vec{b}\) and the squared magnitude of \(\vec{b}\).
Step 3: Detailed Explanation:
We are given the vectors:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k} \] \[ \vec{b} = \hat{i} + 2\hat{j} + \hat{k} \] First, we calculate the dot product \(\vec{a} \cdot \vec{b}\).
\[ \vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) \] \[ \vec{a} \cdot \vec{b} = 2 + 6 + 2 = 10 \] Next, we calculate the magnitude of \(\vec{b}\), denoted as \(|\vec{b}|\).
\[ |\vec{b}| = \sqrt{(1)^2 + (2)^2 + (1)^2} \] \[ |\vec{b}| = \sqrt{1 + 4 + 1} = \sqrt{6} \] Now, we find the square of the magnitude, \(|\vec{b}|^2\).
\[ |\vec{b}|^2 = (\sqrt{6})^2 = 6 \] Finally, we substitute these values into the projection formula.
\[ \text{proj}_{\vec{b}} \vec{a} = \left( \frac{10}{6} \right) (\hat{i} + 2\hat{j} + \hat{k}) \] Simplifying the scalar fraction: \[ \text{proj}_{\vec{b}} \vec{a} = \frac{5}{3} (\hat{i} + 2\hat{j} + \hat{k}) \] We can also distribute the scalar to get the vector components: \[ \text{proj}_{\vec{b}} \vec{a} = \frac{5}{3}\hat{i} + \frac{10}{3}\hat{j} + \frac{5}{3}\hat{k} \] Step 4: Final Answer:
The projection of vector \(\vec{a}\) on vector \(\vec{b}\) is \(\frac{5}{3}(\hat{i} + 2\hat{j} + \hat{k})\).