Worked examples · Sheet 1

Operations with components, magnitudes and unit vectors

Theory used: Operations in components Vector between two points and midpoint Magnitude of a vector Distance between two points Unit vector Linear combination

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Given $u = (- 2, 1, 2)$ and $v = (- 1, 1, - 1)$ , compute the vector $3 u - 2 v$ and its magnitude.

3 u = (- 6, 3, 6), 2 v = (- 2, 2, - 2)

3 u - 2 v = (- 6 - (- 2), 3 - 2, 6 - (- 2)) = (- 4, 1, 8)

∥3 u - 2 v ∥ = (- 4)^{2} + 1^{2} + 8^{2} = 16 + 1 + 64 = 81 = 9

Operations in components are performed one component at a time: multiplication by a scalar multiplies each component by the scalar, and the difference subtracts the corresponding components. One therefore first computes $3 u$ and $2 v$ , then their difference, paying attention to the signs of the negative components of $2 v$ . The magnitude is the square root of the sum of the squares of the components, therefore the vector $(- 4, 1, 8)$ has magnitude $9$ .

Exercise 2

Given the points $A = (1, 2, - 1)$ and $B = (3, - 2, 3)$ , find the vector $A B$ , the distance $d (A, B)$ , the midpoint $M$ of the segment $A B$ and the unit vector of $A B$ .

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

d (A, B) = ∥ A B ∥ = 4 + 16 + 16 = 36 = 6

M = (\frac{1 + 3}{2}, \frac{2 + ( - 2 )}{2}, \frac{- 1 + 3}{2}) = (2, 0, 1)

\frac{A B}{∥ A B ∥} = \frac{1}{6} (2, - 4, 4) = (\frac{1}{3}, - \frac{2}{3}, \frac{2}{3})

The components of the vector $A B$ are the differences between the coordinates of the final point $B$ and those of the initial point $A$ , in that order. The distance between the two points is the magnitude of $A B$ , therefore $d (A, B) = 6$ . The coordinates of the midpoint are the averages of the coordinates of $A$ and of $B$ . The unit vector is obtained by dividing each component of $A B$ by the magnitude $6$ ; the result has magnitude $1$ , since $\frac{1}{9} + \frac{4}{9} + \frac{4}{9} = 1$ , and it has the same direction and the same orientation as $A B$ .

Exercise 3

Find the vectors of magnitude $6$ parallel to $v = (2, - 1, 2)$ and, among them, the one with orientation opposite to $v$ .

∥ v ∥ = 4 + 1 + 4 = 3, \frac{v}{∥ v ∥} = (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})

w = \pm 6 \cdot \frac{v}{∥ v ∥} = \pm (4, - 2, 4)

orientation opposite to v : w = (- 4, 2, - 4) = - 2 v

A vector parallel to $v$ is written as a multiple of the unit vector of $v$ , which has magnitude $1$ ; multiplying the unit vector by $6$ and by $- 6$ gives the two vectors of magnitude $6$ with the direction of $v$ . The positive coefficient preserves the orientation, the negative coefficient reverses it. Therefore the required vectors are $(4, - 2, 4)$ and $(- 4, 2, - 4)$ , and the one with opposite orientation to $v$ is $(- 4, 2, - 4)$ . As a check, $(- 4, 2, - 4) = - 2 v$ and $∥ - 2 v ∥ = 2 \cdot 3 = 6$ .

Exercise 4

Determine whether the vector $w = (5, 2, 8)$ is a linear combination of $u = (1, 1, 1)$ and $v = (1, 0, 2)$ and, if so, find the coefficients.

α (1, 1, 1) + β (1, 0, 2) = (5, 2, 8) ⟺ ⎩ ⎨ ⎧ α + β = 5 α = 2 α + 2 β = 8

α = 2, β = 5 - 2 = 3, α + 2 β = 2 + 6 = 8

w = 2 u + 3 v

Equating the components of the two sides gives a system of three equations in the two unknowns $α$ and $β$ . The second equation immediately gives $α = 2$ , and the first then gives $β = 3$ . With two unknowns and three equations the third equation must be checked: if it were not satisfied, the system would have no solutions and $w$ would not be a linear combination of $u$ and $v$ . Since $2 + 2 \cdot 3 = 8$ , the third equation is satisfied, therefore $w$ is a linear combination of $u$ and $v$ with coefficients $2$ and $3$ .