Question

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

Hint:

The projection of one vector onto another is found by using the formula \( \text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b} \).

Answer 1

Step 1: Formula for projection.
The formula to find the projection of vector \( \mathbf{a} \) on vector \( \mathbf{b} \) is given by: \[ \text{proj}_{\mathbf{b}} \mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}. \]

Step 2: Compute dot product \( \mathbf{a} \cdot \mathbf{b} \).
The dot product of \( \mathbf{a} = \hat{i} + 2 \hat{j} + 3 \hat{k} \) and \( \mathbf{b} = 2 \hat{i} + \hat{k} \) is: \[ \mathbf{a} \cdot \mathbf{b} = (1)(2) + (2)(0) + (3)(1) = 2 + 0 + 3 = 5. \]

Step 3: Compute magnitude of \( \mathbf{b} \).
The magnitude of \( \mathbf{b} \) is: \[ |\mathbf{b}|^2 = (2)^2 + (0)^2 + (1)^2 = 4 + 0 + 1 = 5. \]

Step 4: Find projection.
Now, substitute these values into the projection formula: \[ \text{proj}_{\mathbf{b}} \mathbf{a} = \frac{5}{5} \mathbf{b} = \mathbf{b}. \] Thus, the projection of \( \mathbf{a} \) on \( \mathbf{b} \) is: \[ \text{proj}_{\mathbf{b}} \mathbf{a} = 2 \hat{i} + \hat{k}. \]

Step 5: Conclusion.
Therefore, the projection of \( \mathbf{a} \) on \( \mathbf{b} \) is \( 2 \hat{i} + \hat{k} \).