Orthogonal Vectors

Two vectors are orthogonal (or perpendicular) if the angle between them is a right angle. Even though we cannot use the visual aid of a right angle in dimensions higher than the third dimension, the concept of two β€œorthogonal vectors” still exists in the fourth, fifth, and higher dimensions.

The notion of two vectors v⃗\vec{v} and w⃗\vec{w} being orthogonal is denoted by the symbol

vβƒ—βŠ₯wβƒ—.\vec{v} \perp \vec{w}.

We observe that the length hypotenuse of a right-angled triangle formed by connecting two vectors starting from the same origin is just βˆ₯vβƒ—+wβƒ—βˆ₯\lVert \vec{v} + \vec{w} \rVert:

We remind ourselves of how we calculate the length of a vector:

βˆ₯vβƒ—βˆ₯=vβƒ—β‹…vβƒ—.\lVert \vec{v} \rVert = \sqrt{\vec{v} \cdot \vec{v}}.

We make use of the Pythagorean Theorem:

βˆ₯vβƒ—βˆ₯2+βˆ₯wβƒ—βˆ₯2=βˆ₯vβƒ—+wβƒ—βˆ₯2.\lVert \vec{v} \rVert^2 + \lVert \vec{w} \rVert^2 = \lVert \vec{v} + \vec{w} \rVert^2.

We simplify:

0=βˆ₯vβƒ—βˆ₯2+βˆ₯wβƒ—βˆ₯2βˆ’βˆ₯vβƒ—+wβƒ—βˆ₯2=vβƒ—β‹…vβƒ—+wβƒ—β‹…wβƒ—βˆ’(vβƒ—+wβƒ—)β‹…(vβƒ—+wβƒ—)=vβƒ—β‹…vβƒ—+wβƒ—β‹…wβƒ—βˆ’(vβƒ—β‹…vβƒ—+2vβƒ—β‹…wβƒ—+wβƒ—β‹…wβƒ—)=vβƒ—β‹…vβƒ—+wβƒ—β‹…wβƒ—βˆ’vβƒ—β‹…vβƒ—βˆ’2vβƒ—β‹…wβƒ—βˆ’wβƒ—β‹…wβƒ—=βˆ’2vβƒ—β‹…wβƒ—,\begin{align*} 0 &= \lVert \vec{v} \rVert^2 + \lVert \vec{w} \rVert^2 - \lVert \vec{v} + \vec{w} \rVert^2 \\ &= \vec{v} \cdot \vec{v} + \vec{w} \cdot \vec{w} - (\vec{v} + \vec{w}) \cdot (\vec{v} + \vec{w}) \\ &= \vec{v} \cdot \vec{v} + \vec{w} \cdot \vec{w} - (\vec{v} \cdot \vec{v} + 2\vec{v} \cdot \vec{w} + \vec{w} \cdot \vec{w}) \\ &= \vec{v} \cdot \vec{v} + \vec{w} \cdot \vec{w} - \vec{v} \cdot \vec{v} - 2\vec{v} \cdot \vec{w} - \vec{w} \cdot \vec{w} \\ &= - 2\vec{v} \cdot \vec{w}, \\ \end{align*}

and therefore

v⃗⋅w⃗=0.\vec{v} \cdot \vec{w} = 0.

What have we just shown? The Pythagorean Theorem - which remember, only holds for right-angled triangles - holds if and only if the dot product of the two vectors that we used to form the triangle with equals 00.

Therefore, if the dot product between the two vectors is 00, the triangle they form must be a right-angled triangle, which means that the two vectors are orthogonal.

The notion of a β€œright angle” is not available in higher dimensions, which is why - in higher dimensions - orthogonality is defined through the dot product definition shown above.

So in practice, if the dot product between two real vectors is 00, they are orthogonal. If it isn’t 00, they’re not. This holds for any number of dimensions.

Zero Vector

According to this definition, the zero vector is orthogonal to all other vectors, since for all vectors v⃗\vec{v}

v⃗⋅0⃗=0.\vec{v} \cdot \vec{0} = 0.

Examples

The two vectors

v⃗=[123]\vec{v} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}

and

wβƒ—=[2βˆ’10]\vec{w} = \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}

are orthogonal, because

vβƒ—β‹…wβƒ—=[123]β‹…[2βˆ’10]=1β‹…2+2β‹…(βˆ’1)+3β‹…0=2βˆ’2+0=0.\begin{align*} \vec{v} \cdot \vec{w} &= \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \cdot \begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix} \\ &= 1 \cdot 2 + 2 \cdot (-1) + 3 \cdot 0 \\ &= 2 - 2 + 0 \\ &= 0. \end{align*}

On the other hand, the two vectors

v⃗=[38]\vec{v} = \begin{bmatrix} 3 \\ 8 \end{bmatrix}

and

wβƒ—=[βˆ’60]\vec{w} = \begin{bmatrix} -6 \\ 0 \end{bmatrix}

are not orthogonal, because

vβƒ—β‹…wβƒ—=[38]β‹…[βˆ’60]=3β‹…(βˆ’6)+8β‹…0=βˆ’18+0=βˆ’18β‰ 0.\begin{align*} \vec{v} \cdot \vec{w} &= \begin{bmatrix} 3 \\ 8 \end{bmatrix} \cdot \begin{bmatrix} -6 \\ 0 \end{bmatrix} \\ &= 3 \cdot (-6) + 8 \cdot 0 \\ &= -18 + 0 \\ &= -18 \ne 0. \end{align*}