Question

Orthogonal projection of $\vec{v}$ on $\vec{a}$ is

• A. $\frac{(\vec{a} \cdot \vec{v})}{\|\vec{a}\|} \vec{v}$
• B. $(\vec{a} \cdot \vec{v}) \vec{a}$
• C. $\frac{(\vec{a} \cdot \vec{v})}{\|\vec{a}\|} \vec{a}$
• D. $\frac{(\vec{a} \cdot \vec{v})}{\|\vec{a}\|^2} \vec{a}$

Hint:

Check units/dimensions: $\vec{a} \cdot \vec{v}$ has units of $\|\vec{a}\| \|\vec{v}\|$. Dividing by $\|\vec{a}\|^2$ leaves a pure scalar multiplier for vector $\vec{a}$. Option (D) is the only dimensionally consistent formula!

Answer 1

The Correct Option is D

Step 1: Concept
The vector projection (orthogonal projection) of a vector $\vec{v}$ onto a non-zero vector $\vec{a}$ is a vector parallel to $\vec{a}$ that represents the component of $\vec{v}$ lying along the direction of $\vec{a}$.

Step 2: Key Formulas and Approach
1. The scalar projection (component) of $\vec{v}$ along $\vec{a}$ is given by: \[ \text{comp}_{\vec{a}} \vec{v} = \frac{\vec{a} \cdot \vec{v}}{\|\vec{a}\|} \] 2. The unit vector in the direction of $\vec{a}$ is: \[ \hat{u}_{\vec{a}} = \frac{\vec{a}}{\|\vec{a}\|} \] 3. Multiply the scalar projection by the unit vector to obtain the vector projection.

Step 3: Step-by-step Explanation

• Construct the orthogonal vector projection formula: \[ \text{proj}_{\vec{a}} \vec{v} = (\text{comp}_{\vec{a}} \vec{v}) \, \hat{u}_{\vec{a}} \] \[ = \left( \frac{\vec{a} \cdot \vec{v}}{\|\vec{a}\|} \right) \left( \frac{\vec{a}}{\|\vec{a}\|} \right) \] \[ = \frac{\vec{a} \cdot \vec{v}}{\|\vec{a}\|^2} \, \vec{a} \]
• Notice that $\|\vec{a}\|^2 = \vec{a} \cdot \vec{a}$, so it can also be expressed as $\frac{(\vec{a} \cdot \vec{v})}{(\vec{a} \cdot \vec{a})} \vec{a}$.

Step 4: Final Answer
The orthogonal projection of $\vec{v}$ on $\vec{a}$ is $\frac{(\vec{a} \cdot \vec{v})}{\|\vec{a}\|^2} \vec{a}$. Thus, Option (D) is correct.