Worked examples · Sheet 3

Orthogonal vectors and orthogonal projections

Theory used: Orthogonal vectors Dot product in components Orthogonal projection onto a vector Properties of the dot product

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Find the values of $k \in R$ for which the vectors $u = (k, 2, - 1)$ and $v = (k, k, 3)$ are orthogonal.

u \cdot v = k \cdot k + 2 \cdot k + (- 1) \cdot 3 = k^{2} + 2 k - 3

k^{2} + 2 k - 3 = 0 ⟺ (k + 3) (k - 1) = 0 ⟺ k = - 3 or k = 1

Two vectors are orthogonal when their dot product is zero, so the condition translates into an equation in the parameter. The dot product in components is the polynomial $k^{2} + 2 k - 3$ , which factors as $(k + 3) (k - 1)$ . Therefore the vectors are orthogonal for $k = - 3$ and for $k = 1$ . For $k = 1$ , for example, one has $u = (1, 2, - 1)$ , $v = (1, 1, 3)$ and $u \cdot v = 1 + 2 - 3 = 0$ .

Exercise 2

Decompose the vector $v = (4, 3)$ as the sum of a vector parallel to $u = (1, 2)$ and a vector orthogonal to $u$ .

v \cdot u = 4 + 6 = 10, ∥ u ∥^{2} = 1 + 4 = 5

v_{∥} = \frac{v \cdot u}{∥ u ∥ ^{2}} u = \frac{10}{5} (1, 2) = (2, 4)

v_{⊥} = v - v_{∥} = (4, 3) - (2, 4) = (2, - 1)

v_{⊥} \cdot u = 2 - 2 = 0

v = (4, 3)

is the sum of

v_{∥} = (2, 4)

, on the line of

u

(dashed), and

v_{⊥} = (2, - 1)

, orthogonal to that line

The required decomposition is the one in the theorem on the orthogonal projection: the parallel part is the projection of $v$ onto the direction of $u$ , the orthogonal part is the difference between $v$ and the projection. The coefficient $\frac{v \cdot u}{∥ u ∥ ^{2}}$ equals $2$ , therefore $v_{∥} = 2 u = (2, 4)$ and $v_{⊥} = (2, - 1)$ . The dot product $v_{⊥} \cdot u$ is zero, therefore the orthogonal part is indeed orthogonal to $u$ , and the sum of the two parts gives back $v$ .

Exercise 3

Compute the orthogonal projection of $v = (3, 4, - 4)$ onto the direction of $u = (2, 1, - 2)$ and the component of $v$ orthogonal to $u$ .

v \cdot u = 6 + 4 + 8 = 18, ∥ u ∥^{2} = 4 + 1 + 4 = 9

pr_{u} (v) = \frac{18}{9} (2, 1, - 2) = (4, 2, - 4)

v_{⊥} = (3, 4, - 4) - (4, 2, - 4) = (- 1, 2, 0), v_{⊥} \cdot u = - 2 + 2 + 0 = 0

The procedure is the same as in the plane, with three components. The dot product $v \cdot u$ equals $18$ and the square of the magnitude of $u$ equals $9$ , so the coefficient of the projection is $2$ and the orthogonal projection is $(4, 2, - 4)$ . Subtracting the projection from $v$ gives the orthogonal component $(- 1, 2, 0)$ , and the zero dot product with $u$ confirms the computation. The number $\frac{v \cdot u}{∥ u ∥} = \frac{18}{3} = 6$ is the signed length of the projection, positive because the angle between $v$ and $u$ is acute.

Exercise 4

Given the triangle with vertices $A = (0, 1, 1)$ , $B = (4, - 3, 3)$ , $C = (1, - 2, 2)$ , find the foot $H$ of the altitude from $C$ and the length of the altitude $C H$ .

A B = (4, - 4, 2), A C = (1, - 3, 1)

A C \cdot A B = 4 + 12 + 2 = 18, ∥ A B ∥^{2} = 16 + 16 + 4 = 36

A H = pr_{A B} (A C) = \frac{18}{36} (4, - 4, 2) = (2, - 2, 1)

H = A + A H = (0 + 2, 1 - 2, 1 + 1) = (2, - 1, 2)

H C = A C - A H = (1, - 3, 1) - (2, - 2, 1) = (- 1, - 1, 0), ∥ H C ∥ = 2

The foot $H$ of the altitude is the point of the line $A B$ such that $H C$ is orthogonal to $A B$ . Applying the vector $A C$ at the point $A$ and decomposing it with respect to the direction of $A B$ , the parallel part is $A H$ and the orthogonal part is $H C$ . The coefficient of the projection equals $\frac{18}{36} = \frac{1}{2}$ , therefore $H$ is the midpoint of the side $A B$ , that is, $H = (2, - 1, 2)$ . The altitude is the magnitude of the orthogonal part, therefore $C H = 2$ . As a check, $H C \cdot A B = - 4 + 4 + 0 = 0$ .