Calculus Orthonormal

Question

Find an orthonormal basis of the plane x1−4x2−x3=0.

Answer 1

We are tasked with finding an orthonormal basis for the plane defined by the equation:

x1−4x2−x3=0.x_1 - 4x_2 - x_3 = 0.Step 1: Understand the plane

This equation defines a plane in R3\mathbb{R}^3, and any vector lying in this plane satisfies the equation. To find an orthonormal basis, we:

1. Identify a general form for the plane.
2. Find two linearly independent vectors in the plane.
3. Use the Gram-Schmidt process to make the vectors orthogonal.
4. Normalize the orthogonal vectors to make them unit vectors.

Step 2: Parametrize the plane

Rewriting the equation:

x1=4x2+x3.x_1 = 4x_2 + x_3.Let x2=tx_2 = t and x3=sx_3 = s (free parameters). Then:

x=[x1x2x3]=[4t+sts].\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 4t + s \\ t \\ s \end{bmatrix}.Expressing x\mathbf{x} as a linear combination of tt and ss:

x=t[410]+s[101].\mathbf{x} = t \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} + s \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.Thus, two linearly independent vectors in the plane are:

v1=[410],v2=[101].\mathbf{v}_1 = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}. Step 3: Gram-Schmidt orthogonalization

To make these vectors orthogonal, we use the Gram-Schmidt process:

Step 3.1: Keep u1=v1\mathbf{u}_1 = \mathbf{v}_1:

u1=[410].\mathbf{u}_1 = \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix}.Step 3.2: Subtract the projection of v2\mathbf{v}_2 onto u1\mathbf{u}_1:

The projection formula is:

proju1v2=v2⋅u1u1⋅u1u1.\text{proj}_{\mathbf{u}_1} \mathbf{v}_2 = \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \mathbf{u}_1.Compute the dot products:

v2⋅u1=4(1)+1(0)+0(1)=4,\mathbf{v}_2 \cdot \mathbf{u}_1 = 4(1) + 1(0) + 0(1) = 4, u1⋅u1=42+12+02=16+1=17.\mathbf{u}_1 \cdot \mathbf{u}_1 = 4^2 + 1^2 + 0^2 = 16 + 1 = 17.So:

proju1v2=417[410]=[16174170].\text{proj}_{\mathbf{u}_1} \mathbf{v}_2 = \frac{4}{17} \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} \frac{16}{17} \\ \frac{4}{17} \\ 0 \end{bmatrix}.Now subtract this projection from v2\mathbf{v}_2 to find u2\mathbf{u}_2:

u2=v2−proju1v2=[101]−[16174170].\mathbf{u}_2 = \mathbf{v}_2 - \text{proj}_{\mathbf{u}_1} \mathbf{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} - \begin{bmatrix} \frac{16}{17} \\ \frac{4}{17} \\ 0 \end{bmatrix}. u2=[1−16170−4171−0]=[117−4171].\mathbf{u}_2 = \begin{bmatrix} 1 - \frac{16}{17} \\ 0 - \frac{4}{17} \\ 1 - 0 \end{bmatrix} = \begin{bmatrix} \frac{1}{17} \\ -\frac{4}{17} \\ 1 \end{bmatrix}. Step 4: Normalize the vectors

To make the vectors orthonormal, divide each by its magnitude.

Normalize u1\mathbf{u}_1:

∥u1∥=42+12+02=17.\|\mathbf{u}_1\| = \sqrt{4^2 + 1^2 + 0^2} = \sqrt{17}. u1=117[410].\mathbf{u}_1 = \frac{1}{\sqrt{17}} \begin{bmatrix} 4 \\ 1 \\ 0 \end{bmatrix}.Normalize u2\mathbf{u}_2:

∥u2∥=(117)2+(−417)2+12.\|\mathbf{u}_2\| = \sqrt{\left(\frac{1}{17}\right)^2 + \left(-\frac{4}{17}\right)^2 + 1^2}. ∥u2∥=1289+16289+1=17289+1=306289=30617.\|\mathbf{u}_2\| = \sqrt{\frac{1}{289} + \frac{16}{289} + 1} = \sqrt{\frac{17}{289} + 1} = \sqrt{\frac{306}{289}} = \frac{\sqrt{306}}{17}. u2=1∥u2∥[117−4171]=17306[117−4171].\mathbf{u}_2 = \frac{1}{\|\mathbf{u}_2\|} \begin{bmatrix} \frac{1}{17} \\ -\frac{4}{17} \\ 1 \end{bmatrix} = \frac{17}{\sqrt{306}} \begin{bmatrix} \frac{1}{17} \\ -\frac{4}{17} \\ 1 \end{bmatrix}. u2=1306[1−417].\mathbf{u}_2 = \frac{1}{\sqrt{306}} \begin{bmatrix} 1 \\ -4 \\ 17 \end{bmatrix}. Final Orthonormal Basis:

Final Answer:

u1​=1/sqrt 17​ [410]​​,u2​=1/sqrt 306 ​[1−417​​]