Vectors

Solution to Exercise 3.1

(a)   \(2\mathbf{u} + \mathbf{w} = 2\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \begin{pmatrix} 1 \\ 6 \end{pmatrix} = \begin{pmatrix} 4 \\ 6 \end{pmatrix} + \begin{pmatrix} 1 \\ 6 \end{pmatrix} = \begin{pmatrix} 5 \\ 12 \end{pmatrix}\)

(b)   \(\mathbf{w} - \mathbf{u} = \begin{pmatrix} 1 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 1 - 2 \\ 6 - 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

(c)   \(\hat{\mathbf{u}} = \dfrac{\mathbf{u}}{\|\mathbf{u}\|} = \dfrac{1}{\sqrt{13}} \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{2}{\sqrt{13}} \\ \frac{3}{\sqrt{13}} \end{pmatrix}\)

(d)   \(-\hat{\mathbf{v}} = -\dfrac{\mathbf{v}}{\|\mathbf{v}\|} = -\dfrac{1}{\sqrt{13}} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} -\frac{3}{\sqrt{13}} \\ \frac{2}{\sqrt{13}} \end{pmatrix}\)

(e)   \(\dfrac{1}{2}\mathbf{v} = \dfrac{1}{2} \begin{pmatrix} 3 \\ -2 \end{pmatrix} = \begin{pmatrix} 3 / 2 \\ -2 / 2 \end{pmatrix}\)

(f)   \(\mathbf{v} - \mathbf{u} = \begin{pmatrix} 3 \\ -2 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix}= \begin{pmatrix} 1 \\ -5 \end{pmatrix}\)

(g)   \(\mathbf{w} - \mathbf{u} = \begin{pmatrix} 1 \\ 6 \end{pmatrix} - \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} -1 \\ 3 \end{pmatrix}\)

(h)   \(\mathbf{u} \cdot \mathbf{w} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 6 \end{pmatrix} = 2 \times 1 + 3 \times 6 = 20\)

(i)   Using equation (3.4)

\[\begin{split} \begin{align*} (\mathbf{v} - \mathbf{u}) \cdot (\mathbf{w} - \mathbf{u}) &= \|(\mathbf{v} - \mathbf{u})\|\|(\mathbf{w} - \mathbf{u})\| \cos(\theta) \\ \begin{pmatrix} 1 \\ -5 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 3 \end{pmatrix} &= \left\| \begin{pmatrix} 1 \\ -5 \end{pmatrix} \right\| \left\| \begin{pmatrix} -1 \\ 3 \end{pmatrix} \right\| \cos(\theta) \\ -16 &= \sqrt{26} \sqrt{10} \cos(\theta) \\ \theta &= \cos^{-1} \left( \frac{-16}{\sqrt{260}} \right) \approx 3.0172 \end{align*} \end{split}\]

(j)   \(\mathbf{u} \cdot \mathbf{v} = \begin{pmatrix} 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -2 \end{pmatrix} = 2 \times 3 + 3 \times (-2) = 0\)

(k)   \(\mathbf{v} \times \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -2 & 0 \\ 1 & 6 & 0 \end{vmatrix} = 0\mathbf{i} - 0 \mathbf{j} + 20 \mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 20 \end{pmatrix}\)

Solution to Exercise 3.2

(a)   \(\mathbf{u} = 2 \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} + 7 \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = 2 \mathbf{i} + 7 \mathbf{j} + \mathbf{k}\)

(b)

\[\begin{split} \begin{align*} & \left( \begin{array}{ccc\|c} 1 & 0 & 1 & 2 \\ -1 & 2 & 0 & 7 \\ 0 & 0 & -1 & 1 \end{array} \right) \begin{array}{l} \\ R_2 + R_1 \\ \phantom{x} \end{array} & \longrightarrow & \left( \begin{array}{ccc\|c} 1 & 0 & 1 & 2 \\ 0 & 2 & 1 & 9 \\ 0 & 0 & -1 & 1 \end{array} \right) \begin{array}{l} \\ \frac{1}{2} R_2 \\ -R_3 \phantom{x} \end{array} \\ \\ \longrightarrow & \left( \begin{array}{ccc\|c} 1 & 0 & 1 & 2 \\ 0 & 1 & 1/2 & 9/2 \\ 0 & 0 & 1 & -1 \end{array} \right) \begin{array}{l} R_1 - R_3 \\ R_2 - \frac{1}{2} R_3 \\ \phantom{x} \end{array} & \longrightarrow & \left( \begin{array}{ccc\|c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 5 \\ 0 & 0 & 1 & -1 \end{array} \right) \begin{array}{l} R_1 - R_3 \\ R_2 - \frac{1}{2} R_3 \\ \phantom{x} \end{array} \\ \\ \end{align*} \end{split}\]

Therefore \(\mathbf{u} = 3 \mathbf{f}_1 + 5 \mathbf{f}_2 - \mathbf{f}_3\).

Solution to Exercise 3.3

(a)   If \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular then \(\mathbf{u} \cdot \mathbf{v} = 0\).

\[\begin{split} \begin{align*} \mathbf{u} \cdot \mathbf{v} &= 0 \\ \begin{pmatrix} 1 \\ k \\ -2 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -5 \\ 4 \end{pmatrix} &= 0 \\ 2 - 5 k - 8 &= 0 \\ -5k &= 6 \\ \therefore k &= -\frac{6}{5}. \end{align*} \end{split}\]

(b)  

\[\begin{split} \begin{align*} \mathbf{u} \cdot \mathbf{v} &= 0 \\ \begin{pmatrix} 1 \\ 0 \\ k + 2 \\ -1 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ k \\ -2 \\ 1 \\ 2 \end{pmatrix} &= 0 \\ 1 - 2k - 2 - 1 + 2 &= 0 \\ -2k &= 0 \\ \therefore k &= 0. \end{align*} \end{split}\]

Solution to Exercise 3.4

\[\begin{split} \begin{align*} \mathbf{u} \cdot \mathbf{v} &= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix} = -1 + 4 - 3 = 0, \\ \mathbf{u} \cdot \mathbf{w} &= \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 2 - 6 + 3 = -1, \\ \mathbf{v} \cdot \mathbf{w} &= \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = -2 -6 -1 = -9. \end{align*} \end{split}\]

Therefore \(\mathbf{u} \perp \mathbf{v}\). The angle between \(\mathbf{u}\) and \(\mathbf{w}\) is

\[\begin{split} \begin{align*} \theta &= \cos^{-1} \left( \frac{\mathbf{u} \cdot \mathbf{w}}{\|\mathbf{u} \|\|\mathbf{w}\|} \right) \\ &= \cos^{-1} \left( \frac{-1}{\sqrt{14}\sqrt{14}}\right) \\ &= \cos^{-1}\left(\frac{-1}{14}\right) \approx 1.6423, \end{align*} \end{split}\]

and the angle between \(\mathbf{v}\) and \(\mathbf{w}\) is

\[\begin{split} \begin{align*} \theta &= \cos^{-1} \left( \frac{\mathbf{v} \cdot \mathbf{w}}{\|\mathbf{v} \|\|\mathbf{w}\|} \right) \\ &= \cos^{-1} \left( \frac{-9}{\sqrt{6}\sqrt{14}} \right) \approx 2.9515. \end{align*} \end{split}\]