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Dot Product Properties

Orthogonal Vectors

Vectors u⃗\vec{u} and v⃗\vec{v} are orthogonal or perpendicular if and only if u⃗⋅v⃗=0\vec{u}\cdot \vec{v}=0.


Shortcut in R2\colorThree{\reals^2}

Given a vector [ab]∈R2\begin{bmatrix} a\\ b\\ \end{bmatrix} \in \reals^2, we can "switch and flip" to find two non-trivial orthogonal vectors: [b−a]\begin{bmatrix} b\\ -a\\ \end{bmatrix} and [−ba]\begin{bmatrix} -b\\ a\\ \end{bmatrix} .

Wize Concept
Non-trivial usually means non-zero: the zero vector is orthogonal to all vectors since [v1v2⋮vn]⋅[00⋮0]=0\begin{bmatrix} v_1\\ v_2\\ \vdots\\ v_n \end{bmatrix} \cdot \begin{bmatrix} 0\\ 0\\ \vdots\\ 0 \end{bmatrix} =0.

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Properties

Let u⃗, v⃗, w⃗ ∈Rn\vec{u},\ \vec{v},\ \vec{w} \ \in \reals^n be vectors.
Let c∈Rc \in \reals be a scalar.
  • u⃗⋅v⃗=v⃗⋅u⃗\vec{u}\cdot\vec{v}=\vec{v}\cdot\vec{u}
  • v⃗⋅0⃗ =0\vec{v}\cdot\vec{0}\ =0
  • u⃗⋅u⃗=∣∣u⃗∣∣2\vec{u}\cdot\vec{u}=\left|\left|\vec{u}\right|\right|^2
  • c(u⃗⋅v⃗) = (cu⃗)⋅v⃗ = u⃗⋅(cv⃗)c\left(\vec{u}\cdot\vec{v}\right) \ = \ \left(c\vec{u}\right)\cdot\vec{v} \ =\ \vec{u}\cdot\left(c\vec{v}\right)
  • u⃗⋅(v⃗+w⃗) = u⃗⋅v⃗ + u⃗⋅w⃗\vec{u}\cdot\left(\vec{v}+\vec{w}\right)\ =\ \vec{u}\cdot\vec{v}\ +\ \vec{u}\cdot\vec{w}

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Example: Orthogonal Vectors

Find all values of kk such that the vectors u⃗=[k1]\vec{u}= \begin{bmatrix} k\\ 1\\ \end{bmatrix} and v⃗=[k−1]\vec{v}=\begin{bmatrix} k\\ -1\\ \end{bmatrix} are orthogonal.

u⃗⋅v⃗=0[k1]⋅[k−1]=0k2−1=0k2=1k=±1\begin{aligned} \vec{u} \cdot \vec{v}&=0\\ \begin{bmatrix} k\\ 1\\ \end{bmatrix} \cdot \begin{bmatrix} k\\ -1\\ \end{bmatrix} &= 0\\ k^2 - 1 &= 0\\ k^2 &= 1\\ k &= \boxed{\pm 1} \end{aligned}

Watch Out!
When solving by taking the square root of a number, you must include both ±\pm possibilities.

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Example: Dot Product Properties (Proof)

Prove that if ∥u⃗∥=∥v⃗∥\|\vec{u}\|=\|\vec{v}\| then u⃗+v⃗\vec{u}+\vec{v} is orthogonal to u⃗−v⃗\vec{u}-\vec{v}.

Two vectors are orthogonal if their dot product is 0, so let's find the dot product of u⃗+v⃗\vec{u}+\vec{v} and u⃗−v⃗\vec{u}-\vec{v}:

(u⃗+v⃗)⋅(u⃗−v⃗)=u⃗⋅(u⃗−v⃗)+v⃗⋅(u⃗−v⃗)=u⃗⋅u⃗−u⃗⋅v⃗+v⃗⋅u⃗−v⃗⋅v⃗=u⃗⋅u⃗−u⃗⋅v⃗+u⃗⋅v⃗−v⃗⋅v⃗=∥u⃗∥2−∥v⃗∥2=∥u⃗∥2−∥u⃗∥2since we assume ∥u⃗∥=∥v⃗∥=0\begin{aligned} \colorOne{(\vec{u}+\vec{v})} \cdot (\vec{u}-\vec{v}) &=\colorOne{\vec u} \cdot (\vec{u}-\vec{v}) + \colorOne{\vec{v}} \cdot (\vec{u}-\vec{v})\\ &=\colorOne{\vec{u}} \cdot \vec{u} - \colorOne{\vec{u}} \cdot \vec{v} + \colorOne{\vec{v}} \cdot \vec{u} - \colorOne{\vec{v}} \cdot \vec{v}\\ &=\colorOne{\vec{u}} \cdot \vec{u} - \cancel{\colorOne{\vec{u}} \cdot \vec{v}} + \cancel{\vec{u} \cdot \colorOne{\vec{v}}} - \colorOne{\vec{v}} \cdot \vec{v}\\ &= {\lVert \vec u \rVert}^2 - {\lVert \vec v \rVert}^2\\ &= {\lVert \vec u \rVert}^2 - \colorTwo{{\lVert \vec u \rVert}}^2 \qquad \text{since we assume } \lVert \vec u \rVert = \lVert \vec v \rVert\\ &=0 \end{aligned}
Since the dot product (u⃗+v⃗)⋅(u⃗−v⃗)=0(\vec{u}+\vec{v}) \cdot (\vec u - \vec v) = 0 , the vectors u⃗+v⃗\vec u + \vec v and u⃗−v⃗\vec u - \vec v are orthogonal.

Practice: Properties of Dot Product

Given that u⃗=−57v⃗\vec{u}=-\dfrac{5}{7}\vec{v} and u⃗⋅v⃗=−14\vec{u}\cdot\vec{v}=-14, find ∥u⃗∥\lVert \vec u \rVert.

Practice: Dot Product Properties

Let u⃗, v⃗,w⃗∈R3\vec{u},\ \vec{v}, \vec{w} \in \reals^3 be vectors and let c, d∈Rc,\ d \in \reals be scalars. Select all expressions that are valid.
Extra Practice