Worked examples · Sheet 2

Angles between vectors with the dot product

Theory used: Angle between two vectors Dot product Dot product in components Computing the angle

Exercise 1

Compute the angle between the vectors $u = (1, 0, 1)$ and $v = (- 2, 2, 0)$ .

u \cdot v = 1 \cdot (- 2) + 0 \cdot 2 + 1 \cdot 0 = - 2

∥ u ∥ = 1 + 0 + 1 = 2, ∥ v ∥ = 4 + 4 + 0 = 22

cos θ = \frac{u \cdot v}{∥ u ∥ ∥ v ∥} = \frac{- 2}{2 \cdot 2 2} = \frac{- 2}{4} = - \frac{1}{2} ⟹ θ = \frac{2 π}{3}

The dot product in components is the sum of the products of the corresponding components and equals $- 2$ . The negative sign already indicates that the angle is obtuse. Dividing by the product of the magnitudes gives $cos θ = - \frac{1}{2}$ . Since the angle between two vectors belongs to the interval $[0, π]$ , where cosine is invertible, the only angle with this cosine is $θ = \frac{2 π}{3}$ .

Exercise 2

Compute the angles of the triangle with vertices $A = (1, 0, 1)$ , $B = (2, 1, 1)$ , $C = (0, 0, 2)$ .

A B = (1, 1, 0), A C = (- 1, 0, 1), B C = (- 2, - 1, 1)

cos \hat{A} = \frac{A B \cdot A C}{∥ A B ∥ ∥ A C ∥} = \frac{- 1 + 0 + 0}{2 \cdot 2} = - \frac{1}{2} ⟹ \hat{A} = \frac{2 π}{3}

B A = (- 1, - 1, 0), cos \hat{B} = \frac{B A \cdot B C}{∥ B A ∥ ∥ B C ∥} = \frac{2 + 1 + 0}{2 \cdot 6} = \frac{3}{12} = \frac{3}{2} ⟹ \hat{B} = \frac{π}{6}

\hat{C} = π - \frac{2 π}{3} - \frac{π}{6} = \frac{π}{6}

The angle at a vertex is the angle between the two vectors leaving that vertex. For the vertex $A$ one uses $A B$ and $A C$ , and obtains $\hat{A} = \frac{2 π}{3}$ . For the vertex $B$ the outgoing vectors are $B A = - A B$ and $B C$ ; using $A B$ instead of $B A$ would change the sign of the dot product and would give the supplementary angle. With the correct vectors one gets $cos \hat{B} = \frac{3}{2}$ , therefore $\hat{B} = \frac{π}{6}$ . The third angle follows from the fact that the sum of the angles of a triangle is $π$ , therefore $\hat{C} = \frac{π}{6}$ ; the triangle is isosceles, in agreement with $∥ A B ∥ = ∥ A C ∥ = 2$ .

Exercise 3

Find the values of $k \in R$ for which the angle between $u = (0, 1, 1)$ and $v = (1, k, 0)$ is $\frac{π}{3}$ .

u \cdot v = k, ∥ u ∥ = 2, ∥ v ∥ = 1 + k^{2}

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

{k \geq 0 4 k^{2} = 2 (1 + k^{2}) ⟺ {k \geq 0 k^{2} = 1 ⟺ k = 1

The condition on the angle translates into the equation $cos θ = \frac{1}{2}$ , with the cosine expressed through the dot product and the magnitudes. The equation $2 k = 2 1 + k^{2}$ contains a square root, whose value is nonnegative; therefore squaring is allowed only together with the condition $k \geq 0$ , and the two conditions must be imposed as a system. The squared equation gives $k^{2} = 1$ , that is, $k = \pm 1$ , and the condition $k \geq 0$ excludes $k = - 1$ . Indeed, for $k = - 1$ one would have $cos θ = - \frac{1}{2}$ , that is, the angle $\frac{2 π}{3}$ . The required value is therefore unique, $k = 1$ .