Worked examples · Sheet 1

Norms, angles and orthogonality conditions between vectors

Theory used: Standard scalar product Norm and distance Angle between two vectors Orthogonal vectors Pythagorean theorem

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Given $u = (2, 1, 1)$ and $v = (1, - 1, 2)$ in $R^{3}$ , compute $⟨ u, v ⟩$ , the norms, the angle between $u$ and $v$ and the distance $d (u, v)$ .

⟨ u, v ⟩ = 2 \cdot 1 + 1 \cdot (- 1) + 1 \cdot 2 = 3

∥ u ∥ = 4 + 1 + 1 = 6, ∥ v ∥ = 1 + 1 + 4 = 6

cos θ = \frac{⟨ u , v ⟩}{∥ u ∥ ∥ v ∥} = \frac{3}{6} = \frac{1}{2} ⟹ θ = \frac{π}{3}

d (u, v) = ∥ u - v ∥ = ∥ (1, 2, - 1) ∥ = 1 + 4 + 1 = 6

The standard scalar product is the sum of the products of the components in the same position, and the norm is the square root of the scalar product of a vector with itself. Since $u$ and $v$ are nonzero, the angle is the unique $θ \in [0, π]$ with cosine equal to $\frac{⟨ u , v ⟩}{∥ u ∥ ∥ v ∥}$ ; the quotient equals $\frac{1}{2}$ , therefore $θ = \frac{π}{3}$ . The distance is the norm of the difference. The three sides of the triangle with vertices $0$ , $u$ , $v$ all measure $6$ , so the triangle is equilateral, in agreement with the angle of $\frac{π}{3}$ at the vertex $0$ .

Exercise 2

Determine $k \in R$ so that $u = (k, 1, 2, - 1)$ and $v = (2, k, - 1, 1)$ are orthogonal in $R^{4}$ . For the value found, verify the Pythagorean theorem for $u$ and $v$ .

⟨ u, v ⟩ = 2 k + k - 2 - 1 = 3 k - 3 = 0 ⟹ k = 1

u = (1, 1, 2, - 1), v = (2, 1, - 1, 1), u + v = (3, 2, 1, 0)

∥ u ∥^{2} = 1 + 1 + 4 + 1 = 7, ∥ v ∥^{2} = 4 + 1 + 1 + 1 = 7, ∥ u + v ∥^{2} = 9 + 4 + 1 + 0 = 14

Two vectors are orthogonal if and only if their scalar product is zero. The scalar product of $u$ and $v$ is a polynomial of degree one in $k$ , which vanishes only for $k = 1$ , therefore the vectors are orthogonal for $k = 1$ and for no other value. For $k = 1$ the Pythagorean theorem states that $∥ u + v ∥^{2} = ∥ u ∥^{2} + ∥ v ∥^{2}$ , and the computation confirms it, since $14 = 7 + 7$ .

Exercise 3

Determine the unit vectors of $R^{3}$ orthogonal both to $a = (1, 1, 0)$ and to $b = (1, 0, - 1)$ .

{⟨ x, a ⟩ = x_{1} + x_{2} = 0 ⟨ x, b ⟩ = x_{1} - x_{3} = 0 ⟹ {x_{2} = - x_{1} x_{3} = x_{1}

x = t (1, - 1, 1), t \in R, ∥ (1, - 1, 1) ∥ = 3

x = \frac{1}{3} (1, - 1, 1) or x = - \frac{1}{3} (1, - 1, 1)

A vector $x = (x_{1}, x_{2}, x_{3})$ is orthogonal to $a$ and to $b$ if it makes both scalar products vanish, therefore the two conditions form a homogeneous system of two equations in three unknowns. The two equations are independent, so the solutions depend on one parameter and form the line spanned by $(1, - 1, 1)$ . The unit vectors of this line are the two vectors obtained by dividing $(1, - 1, 1)$ by its norm $3$ and taking both orientations: $\pm \frac{1}{3} (1, - 1, 1)$ .

Exercise 4

Determine $k \in R$ so that the angle between $u = (1, 0, 1)$ and $v = (1, k, 0)$ is $\frac{π}{3}$ .

⟨ u, v ⟩ = 1, ∥ u ∥ = 2, ∥ v ∥ = 1 + k^{2}

\frac{1}{2 1 + k ^{2}} = cos \frac{π}{3} = \frac{1}{2} ⟺ 2 (1 + k^{2}) = 2

2 (1 + k^{2}) = 4 ⟺ k^{2} = 1 ⟺ k = \pm 1

The vector $v$ is nonzero for every $k$ , because its first component is $1$ , therefore the angle is defined for every value of the parameter. Since the cosine function is injective on $[0, π]$ , the angle equals $\frac{π}{3}$ if and only if the cosine of the angle equals $\frac{1}{2}$ . In the equation thus obtained both sides are positive, so squaring produces an equivalent equation, which reduces to $k^{2} = 1$ . The problem therefore has two solutions, $k = 1$ and $k = - 1$ .