Solved problems: Orthogonal projection onto a line and plane

📚 Understanding Orthogonal Projections

Orthogonal projection is a way to find the closest point on a subspace (like a line or a plane) to a given vector. Imagine shining a light directly onto the line or plane; the shadow of the vector is its orthogonal projection.

📏 Orthogonal Projection onto a Line

Let's say we have a vector $\mathbf{v}$ and we want to project it onto a line spanned by a vector $\mathbf{u}$. The formula for the projection of $\mathbf{v}$ onto $\mathbf{u}$ is:

$\text{proj}_{\mathbf{u}}(\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{u}}{\mathbf{u} \cdot \mathbf{u}} \mathbf{u}$

• 🔍 Step 1: Calculate the dot product of $\mathbf{v}$ and $\mathbf{u}$ ($\mathbf{v} \cdot \mathbf{u}$).
• ➗ Step 2: Calculate the dot product of $\mathbf{u}$ with itself ($\mathbf{u} \cdot \mathbf{u}$).
• 🧪 Step 3: Divide the result from Step 1 by the result from Step 2. This gives you a scalar.
• 📈 Step 4: Multiply the scalar from Step 3 by the vector $\mathbf{u}$. This gives you the projection vector.

📐 Example: Projecting onto a Line

Let $\mathbf{v} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ and $\mathbf{u} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Then,

$\mathbf{v} \cdot \mathbf{u} = (2)(1) + (3)(1) = 5$

$\mathbf{u} \cdot \mathbf{u} = (1)(1) + (1)(1) = 2$

$\text{proj}_{\mathbf{u}}(\mathbf{v}) = \frac{5}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5/2 \\ 5/2 \end{bmatrix}$

✈️ Orthogonal Projection onto a Plane

Now, let's consider projecting a vector $\mathbf{v}$ onto a plane defined by two orthogonal vectors $\mathbf{u}_1$ and $\mathbf{u}_2$. The formula is:

$\text{proj}_{\text{plane}}(\mathbf{v}) = \frac{\mathbf{v} \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1 + \frac{\mathbf{v} \cdot \mathbf{u}_2}{\mathbf{u}_2 \cdot \mathbf{u}_2} \mathbf{u}_2$

• 1️⃣ Step 1: Find two orthogonal vectors, $\mathbf{u}_1$ and $\mathbf{u}_2$, that span the plane.
• ➕ Step 2: Project $\mathbf{v}$ onto $\mathbf{u}_1$ and $\mathbf{v}$ onto $\mathbf{u}_2$ using the line projection formula.
• 💡 Step 3: Add the two projection vectors together to get the projection of $\mathbf{v}$ onto the plane.

✨ Example: Projecting onto a Plane

Let $\mathbf{v} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$, $\mathbf{u}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$, and $\mathbf{u}_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$. Then,

$\mathbf{v} \cdot \mathbf{u}_1 = (1)(1) + (2)(0) + (3)(0) = 1$

$\mathbf{u}_1 \cdot \mathbf{u}_1 = (1)(1) + (0)(0) + (0)(0) = 1$

$\mathbf{v} \cdot \mathbf{u}_2 = (1)(0) + (2)(1) + (3)(0) = 2$

$\mathbf{u}_2 \cdot \mathbf{u}_2 = (0)(0) + (1)(1) + (0)(0) = 1$

$\text{proj}_{\text{plane}}(\mathbf{v}) = \frac{1}{1} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \frac{2}{1} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 0 \end{bmatrix}$

✍️ Practice Quiz

Try these practice questions to test your understanding!

1. ❓ Project $\mathbf{v} = \begin{bmatrix} 4 \\ -2 \end{bmatrix}$ onto $\mathbf{u} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}$.
2. ❓ Project $\mathbf{v} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ onto the plane spanned by $\mathbf{u}_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $\mathbf{u}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$.
3. ❓ Project $\mathbf{v} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$ onto $\mathbf{u} = \begin{bmatrix} -1 \\ 3 \end{bmatrix}$.