Orthogonal Vectors in \( \mathbb{R}^n \)

In linear algebra, orthogonality generalizes the concept of perpendicularity to \( \mathbb{R}^n \) vector spaces. Below, orthogonal vectors are formally defined, geometrically interpreted, and demonstrated through step-by-step examples.

Table of Contents

Orthogonal Vectors Definition

Two column vectors \( \textbf{x} \) and \( \textbf{y} \) in \( \mathbb{R}^n \):

\[ \textbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}, \quad \textbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{bmatrix} \]

are orthogonal if and only if their inner product (dot product) \( \textbf{x} \cdot \textbf{y} \) is equal to zero.

Vectors \( \textbf{x} \) and \( \textbf{y} \) are orthogonal \( \iff \textbf{x} \cdot \textbf{y} = x_1 y_1 + x_2 y_2 + \dots + x_n y_n = 0 \)

Properties of Orthogonal Vectors

If \( \textbf{x}, \textbf{y} \), and \( \textbf{z} \) are vectors in \( \mathbb{R}^n \), the following properties hold:

  1. Symmetry: If \( \textbf{x} \) is orthogonal to \( \textbf{y} \), then \( \textbf{y} \) is orthogonal to \( \textbf{x} \).
  2. Zero vector: The zero vector \( \textbf{0} \) is orthogonal to every vector since \( \textbf{0} \cdot \textbf{x} = 0 \).
  3. Scalar multiples: If \( \textbf{x} \) is orthogonal to \( \textbf{y} \), then \( k\textbf{x} \) is orthogonal to \( \textbf{y} \) for any scalar \( k \).
  4. Pythagorean theorem: Vectors \( \textbf{x} \) and \( \textbf{y} \) are orthogonal if and only if \( \|\textbf{x} + \textbf{y}\|^2 = \|\textbf{x}\|^2 + \|\textbf{y}\|^2 \).
  5. Linear independence: A set of non-zero mutually orthogonal vectors is always linearly independent.

Note on Orthonormal Vectors: A set of vectors is orthonormal if the vectors are mutually orthogonal and each vector has a norm (length) of 1.

Geometric Interpretation of Orthogonal Vectors

Below are two 2-dimensional vectors \( \textbf{x} = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \) and \( \textbf{y} = \begin{bmatrix} -2 \\ 3 \end{bmatrix} \). Their inner product is given by \( \textbf{x} \cdot \textbf{y} = (3)(-2) + (2)(3) = -6 + 6 = 0 \), meaning they are orthogonal. Geometrically, they make an angle of \( 90^{\circ} \) or \( \frac{\pi}{2} \) radians.

Two orthogonal vectors in 2D space meeting at a 90 degree angle

Similarly, below are two 3-dimensional vectors \( \textbf{u} = \begin{bmatrix} -2 \\ 2 \\ 3 \end{bmatrix} \) and \( \textbf{v} = \begin{bmatrix} 2 \\ -1 \\ 2 \end{bmatrix} \). Their inner product is \( \textbf{u} \cdot \textbf{v} = (-2)(2) + (2)(-1) + (3)(2) = -4 - 2 + 6 = 0 \), meaning they are also orthogonal and meet at a \( 90^{\circ} \) angle in 3D space.

Two orthogonal vectors in 3D space meeting at a 90 degree angle

Examples With Solutions

Example 1: Testing Orthogonality in \( \mathbb{R}^3 \) and \( \mathbb{R}^4 \)

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a) Are the vectors \( \textbf{x} = \begin{bmatrix} -1 \\ 3 \\ 1 \end{bmatrix} \) and \( \textbf{y} = \begin{bmatrix} -2 \\ 3 \\ -11 \end{bmatrix} \) orthogonal?

b) Are the vectors \( \textbf{u} = \begin{bmatrix} 0 \\ -3 \\ 1 \\ 8 \end{bmatrix} \) and \( \textbf{v} = \begin{bmatrix} 3 \\ 5 \\ -2 \\ 2 \end{bmatrix} \) orthogonal?

Solution

a) Calculate the inner product of vectors \( \textbf{x} \) and \( \textbf{y} \):

\[ \textbf{x} \cdot \textbf{y} = (-1)(-2) + (3)(3) + (1)(-11) = 2 + 9 - 11 = 0 \]

The inner product is equal to zero, therefore vectors \( \textbf{x} \) and \( \textbf{y} \) are orthogonal.

b) Calculate the inner product of vectors \( \textbf{u} \) and \( \textbf{v} \):

\[ \textbf{u} \cdot \textbf{v} = (0)(3) + (-3)(5) + (1)(-2) + (8)(2) = 0 - 15 - 2 + 16 = -1 \]

The inner product is NOT equal to zero (\(-1 \neq 0\)), therefore vectors \( \textbf{u} \) and \( \textbf{v} \) are NOT orthogonal.

Example 2: Finding an Orthogonal Vector using the Cross Product

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Let vectors \( \textbf{u} = \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \) and \( \textbf{v} = \begin{bmatrix} -2 \\ 1 \\ -2 \end{bmatrix} \). Find a vector \( \textbf{w} \) that is orthogonal to both \( \textbf{u} \) and \( \textbf{v} \).

Solution

In 3D space (\(\mathbb{R}^3\)), the cross product of two vectors yields a third vector that is orthogonal to both original vectors. The cross product \( \textbf{w} = \textbf{u} \times \textbf{v} \) is given by the determinant:

\[ \textbf{w} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \\ 1 & 0 & -1 \\ -2 & 1 & -2 \end{vmatrix} = \begin{bmatrix} \begin{vmatrix} 0 & -1 \\ 1 & -2 \end{vmatrix} \\ -\begin{vmatrix} 1 & -1 \\ -2 & -2 \end{vmatrix} \\ \begin{vmatrix} 1 & 0 \\ -2 & 1 \end{vmatrix} \end{bmatrix} \]

Evaluating the \(2 \times 2\) determinants:

  • \( x \)-component: \( (0)(-2) - (-1)(1) = 0 + 1 = 1 \)
  • \( y \)-component: \( -((1)(-2) - (-1)(-2)) = -(-2 - 2) = 4 \)
  • \( z \)-component: \( (1)(1) - (0)(-2) = 1 - 0 = 1 \)
\[ \textbf{w} = \begin{bmatrix} 1 \\ 4 \\ 1 \end{bmatrix} \]

Verification: Check that \( \textbf{w} \) is orthogonal to both \( \textbf{u} \) and \( \textbf{v} \):

\[ \textbf{u} \cdot \textbf{w} = (1)(1) + (0)(4) + (-1)(1) = 1 + 0 - 1 = 0 \] \[ \textbf{v} \cdot \textbf{w} = (-2)(1) + (1)(4) + (-2)(1) = -2 + 4 - 2 = 0 \]

Note: Any scalar multiple of \( \textbf{w} \) (i.e., \( k\textbf{w} \) where \( k \) is a non-zero real number) will also be orthogonal to both \( \textbf{u} \) and \( \textbf{v} \).

Example 3: Solving for Unknown Components

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Find the constants \( a \) and \( b \) such that the vectors \( \textbf{u} = \begin{bmatrix} a \\ 4 \\ -b \end{bmatrix} \) and \( \textbf{v} = \begin{bmatrix} a \\ 1 \\ b \end{bmatrix} \) are orthogonal, given the condition \( a = b + 1 \).

Solution

Because vectors \( \textbf{u} \) and \( \textbf{v} \) are orthogonal, their inner product must be equal to zero:

\[ \textbf{u} \cdot \textbf{v} = (a)(a) + (4)(1) + (-b)(b) = a^2 + 4 - b^2 = 0 \]

Given that \( a = b + 1 \), substitute \( b + 1 \) for \( a \) in the inner product equation:

\[ (b + 1)^2 + 4 - b^2 = 0 \]

Expand the squared binomial and simplify:

\[ (b^2 + 2b + 1) + 4 - b^2 = 0 \] \[ 2b + 5 = 0 \]

Solve for \( b \):

\[ 2b = -5 \implies b = -\frac{5}{2} \]

Now, substitute \( b \) back into the given condition to find \( a \):

\[ a = -\frac{5}{2} + 1 = -\frac{3}{2} \]

Therefore, the vectors are orthogonal when \( a = -\frac{3}{2} \) and \( b = -\frac{5}{2} \).

Verification: \( \textbf{u} \cdot \textbf{v} = \left(-\frac{3}{2}\right)^2 + 4 - \left(-\frac{5}{2}\right)^2 = \frac{9}{4} + 4 - \frac{25}{4} = \frac{9 - 25}{4} + 4 = -4 + 4 = 0 \) ✓

Example 4: Vector Operations and Norms

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Let \( \textbf{u} \), \( \textbf{v} \), and \( \textbf{w} \) be vectors in \( \mathbb{R}^n \) such that:

\[ \textbf{w} = \textbf{v} - 4\alpha\textbf{u} \]

where \( \alpha \) is a scalar. Find \( \alpha \) if \( \textbf{u} \) and \( \textbf{w} \) are orthogonal, the norm of \( \textbf{u} \) is \( 5 \), and the dot product \( \textbf{u}^T \textbf{v} = 3 \).

Solution

Since \( \textbf{u} \) and \( \textbf{w} \) are orthogonal, their inner product is equal to zero:

\[ \textbf{u} \cdot \textbf{w} = 0 \]

Substitute \( \textbf{w} = \textbf{v} - 4\alpha\textbf{u} \) into the equation:

\[ \textbf{u} \cdot (\textbf{v} - 4\alpha\textbf{u}) = 0 \]

Use the distributivity and scalar properties of the inner product to expand the left side:

\[ \textbf{u} \cdot \textbf{v} - 4\alpha(\textbf{u} \cdot \textbf{u}) = 0 \quad \text{(I)} \]

We are given that \( \textbf{u}^T \textbf{v} = 3 \), which is equivalent to \( \textbf{u} \cdot \textbf{v} = 3 \).

The norm (length) of vector \( \textbf{u} \) is defined as \( \|\textbf{u}\| = \sqrt{\textbf{u} \cdot \textbf{u}} \). Since \( \|\textbf{u}\| = 5 \), we have:

\[ \textbf{u} \cdot \textbf{u} = \|\textbf{u}\|^2 = 5^2 = 25 \]

Substitute these values back into equation (I):

\[ 3 - 4\alpha(25) = 0 \implies 3 - 100\alpha = 0 \]

Solve for \( \alpha \):

\[ \alpha = \frac{3}{100} \]

Example 5: Trigonometric Vectors in \( \mathbb{R}^2 \)

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Find all values of \( \theta \) such that the vectors \( \textbf{u} = \begin{bmatrix} \sin(\theta) \\ \cos(\theta) \end{bmatrix} \) and \( \textbf{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} \) are orthogonal.

Solution

For vectors \( \textbf{u} \) and \( \textbf{v} \) to be orthogonal, their inner product must be equal to 0:

\[ \textbf{u} \cdot \textbf{v} = (\sin(\theta))(-1) + (\cos(\theta))(1) = 0 \] \[ -\sin(\theta) + \cos(\theta) = 0 \implies \sin(\theta) = \cos(\theta) \]

Divide both sides by \( \cos(\theta) \) (assuming \( \cos(\theta) \neq 0 \)):

\[ \frac{\sin(\theta)}{\cos(\theta)} = \frac{\cos(\theta)}{\cos(\theta)} \implies \tan(\theta) = 1 \]

Solve for \( \theta \):

\[ \theta = \frac{\pi}{4} + n\pi \]

where \( n \) is any integer.

Check when \( \cos(\theta) = 0 \): If \( \cos(\theta) = 0 \), then \( \sin(\theta) = \pm 1 \), so \( \sin(\theta) \neq \cos(\theta) \), and the original equation \( -\sin(\theta) + \cos(\theta) = 0 \) becomes \( \mp 1 + 0 \neq 0 \). So no additional solutions arise from this case.

Example 6: Mutually Orthogonal Vectors in \( \mathbb{R}^3 \)

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Let \( \textbf{u} = \begin{bmatrix} \sin(\theta)\cos(\phi) \\ \sin(\theta)\sin(\phi) \\ \cos(\theta) \end{bmatrix} \), \( \textbf{v} = \begin{bmatrix} \cos(\theta)\cos(\phi) \\ \cos(\theta)\sin(\phi) \\ -\sin(\theta) \end{bmatrix} \), and \( \textbf{w} = \begin{bmatrix} -\sin(\phi) \\ \cos(\phi) \\ 0 \end{bmatrix} \). Prove that each of these vectors is orthogonal to the other two vectors.

Solution

We must calculate the inner product of each pair of vectors and show they all equal zero.

1) Calculate \( \textbf{u} \cdot \textbf{v} \):

\[ \textbf{u} \cdot \textbf{v} = [\sin(\theta)\cos(\phi)][\cos(\theta)\cos(\phi)] + [\sin(\theta)\sin(\phi)][\cos(\theta)\sin(\phi)] + [\cos(\theta)][-\sin(\theta)] \] \[ \textbf{u} \cdot \textbf{v} = \sin(\theta)\cos(\theta)\cos^2(\phi) + \sin(\theta)\cos(\theta)\sin^2(\phi) - \sin(\theta)\cos(\theta) \]

Factor \( \sin(\theta)\cos(\theta) \) out of the first two terms:

\[ \textbf{u} \cdot \textbf{v} = \sin(\theta)\cos(\theta)[\cos^2(\phi) + \sin^2(\phi)] - \sin(\theta)\cos(\theta) \]

Use the Pythagorean identity \( \cos^2(\phi) + \sin^2(\phi) = 1 \):

\[ \textbf{u} \cdot \textbf{v} = \sin(\theta)\cos(\theta)(1) - \sin(\theta)\cos(\theta) = 0 \]

2) Calculate \( \textbf{u} \cdot \textbf{w} \):

\[ \textbf{u} \cdot \textbf{w} = [\sin(\theta)\cos(\phi)][-\sin(\phi)] + [\sin(\theta)\sin(\phi)][\cos(\phi)] + [\cos(\theta)][0] \] \[ \textbf{u} \cdot \textbf{w} = -\sin(\theta)\cos(\phi)\sin(\phi) + \sin(\theta)\sin(\phi)\cos(\phi) + 0 = 0 \]

3) Calculate \( \textbf{v} \cdot \textbf{w} \):

\[ \textbf{v} \cdot \textbf{w} = [\cos(\theta)\cos(\phi)][-\sin(\phi)] + [\cos(\theta)\sin(\phi)][\cos(\phi)] + [-\sin(\theta)][0] \] \[ \textbf{v} \cdot \textbf{w} = -\cos(\theta)\cos(\phi)\sin(\phi) + \cos(\theta)\sin(\phi)\cos(\phi) + 0 = 0 \]

Because the inner product of every pair evaluates to zero, each vector is mutually orthogonal to the other two.

Geometric Note: These three vectors form an orthonormal basis for \( \mathbb{R}^3 \) when expressed in spherical coordinates, which is why they appear so often in physics and engineering.

Frequently Asked Questions

What does it mean for two vectors to be orthogonal?

Two vectors are orthogonal if and only if their inner product (dot product) equals zero. Geometrically, orthogonal vectors meet at a 90-degree angle.

How do you find a vector orthogonal to two given vectors?

In \( \mathbb{R}^3 \), the cross product of two vectors produces a third vector orthogonal to both. For vectors in higher dimensions, you solve a system of equations where the dot product with each given vector equals zero.

Is the zero vector orthogonal to every vector?

Yes. The zero vector is orthogonal to every vector because the dot product of the zero vector with any vector is always zero.

What is the difference between orthogonal and orthonormal vectors?

Orthogonal vectors are perpendicular to each other (dot product equals zero). Orthonormal vectors are orthogonal AND each has a length (norm) of 1.

Are orthogonal vectors always linearly independent?

Yes, a set of non-zero mutually orthogonal vectors is always linearly independent. However, a set containing the zero vector cannot be linearly independent, even though the zero vector is orthogonal to everything.

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