a)
\[ \left( \begin{array}{c} 15 \\ 60 \end{array} \right) \cdot \left( \begin{array}{c} 35 \\ 5 \end{array} \right) = 15 \cdot 35 + 60 \cdot 5 = 825 \]
b) (i)
\[ d = \sqrt{(15-35)^2 + (60-5)^2} = 58.5 \]
b) (ii) To find the angle between the two position vectors:
\[ \cos(\theta) = \frac{825}{\sqrt{15^2 + 60^2} \cdot \sqrt{35^2 + 5^2}} = 67.84^\circ \]
a)
\[\mathbf{OC} = \left(\begin{array}{c} -150 \\ -60 \end{array}\right) + t\left(\begin{array}{c} 35 \\ 40 \end{array}\right)\]
b) The position vector pointing from the police to the car:
\[\mathbf{PC} = \left(\begin{array}{c} -200 \\ -110 \end{array}\right) + t\left(\begin{array}{c} 35 \\ 40 \end{array}\right)\]
The car is closest, when this vector is perpendicular to the path of the car, hence perpendicular to the cars velocity vector, which we can express using the dot product:
\[\left( \left(\begin{array}{c} -200 \\ -110 \end{array}\right) + t\left(\begin{array}{c} 35 \\ 40 \end{array}\right) \right) \cdot \left(\begin{array}{c} 35 \\ 40 \end{array}\right) = 0\]
\[-7000 + 1225t - 4400 + 1600t = 0\]
\[t = 4.03 \: \text{hours}\]
c) We can find the magnitude of the vector \( \mathbf{PC} \) at the time they are the closest:
\[|\mathbf{PC}| = \sqrt{\left(-200 + 35(4.03)\right)^2 + \left(-110 + 40(4.03)\right)^2} = 78.1 \, \text{km} > 50 \, \text{km}\]
a) The dot product of the 2 direction vectors must be 0.
\[ 0 = \left(\begin{array}{c} k \\ 2k \\ 1 \end{array}\right) \cdot \left(\begin{array}{c} 10 \\ 2k \\ 4 \end{array}\right) \]
\[ 0 = 4k^2 + 10k + 4 \]
\[ 0 = 2(k+2)(2k+1) \]
\[ k = -2\:\:\:\:\: k = -\frac{1}{2} \]
b) When \( k = -2 \),
\[ L_1 : \left(\begin{array}{c} -3 \\ 9 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} -2 \\ -4 \\ 1 \end{array}\right) \]
\[ L_2 : \left(\begin{array}{c} 2 \\ 3 \\ 6 \end{array}\right) + \mu \left(\begin{array}{c} 10 \\ -4 \\ 4 \end{array}\right) \]
Where they intersect, the respective coordinates of the lines must be identical:
\[ -3 - 2\lambda = 2 + 10\mu \]
\[ 9 - 4\lambda = 3 - 4\mu \]
\[ 5 + 1\lambda = 6 + 4\mu \]
Solving the first 2 equations simultaneously we get \( \lambda = \frac{5}{6},\; \mu = -\frac{2}{3} \), we can plug these into the third equation:
\[ 5 + \frac{5}{6} = 6 - 4 \cdot \frac{2}{3} \]
\[ \frac{35}{6} \neq \frac{10}{3} \]
Hence, no constants exist such that these two lines share a point.
a) If they are perpendicular, their dot product is 0.
\[ 0 = \left(\begin{array}{c} 17 \\ k \\ 3k \end{array}\right) \cdot \left(\begin{array}{c} 2 \\ 5 \\ 4 \end{array}\right) \]
\[ 0 = 34 + 5k + 12k = 34 + 17k \]
\[ k = -2 \]
b)
\[ \mathbf{F} = a\left(\begin{array}{c} 17 \\ -2 \\ -6 \end{array}\right) \times \left(\begin{array}{c} 2 \\ 5 \\ 4 \end{array}\right) = a\left(\begin{array}{c} 22 \\ -80 \\ 89 \end{array}\right) \]
\[ |\mathbf{F}| = a\sqrt{22^2 + 80^2 + 89^2} \approx 122a = 61 \]
\[ a = 2 \]
a) The position vector of A is:
\[ \mathbf{OA} = \left(\begin{array}{c} 2 + 8\lambda \\ -5 + 3\lambda \\ 3 + 4\lambda \end{array}\right) \]
Recognize that when A is at this closest point, the line drawn from A to the origin will be perpendicular to the line \(\mathbf{r}\). Hence,
\[ 0 = \left(\begin{array}{c} 2 + 8\lambda \\ -5 + 3\lambda \\ 3 + 4\lambda \end{array}\right) \cdot \left(\begin{array}{c} 8 \\ 3 \\ 4 \end{array}\right) \]
\[ 0 = 13 + 89\lambda \]
\[ \lambda = -\frac{13}{89} \]
Now that we know \(\lambda\), we can plug it into the equation of \(\mathbf{r}\).
\[ \mathbf{r} = \left(\begin{array}{c} \frac{74}{89} \\ -\frac{484}{89} \\ \frac{215}{89} \end{array}\right) \]
a) The vector \(\overrightarrow{\text{XY}} = \left(\begin{array}{c} -1 \\ -1 \\ -1 \end{array}\right)\)
b) The vector \(\overrightarrow{\text{XZ}} = \left(\begin{array}{c} 0 \\ -1 \\ 2 \end{array}\right)\)
c) The cross product:
\[ \overrightarrow{\text{XY}} \times \overrightarrow{\text{XZ}} = \left(\begin{array}{c} -3 \\ 2 \\ 1 \end{array}\right) \]
The area is:
\[ \text{Area} = \frac{1}{2} \left|\overrightarrow{\text{XY}} \times \overrightarrow{\text{XZ}}\right| = \frac{1}{2} \sqrt{9 + 4 + 1} = \frac{\sqrt{14}}{2} \]
\[ |\mathbf{w}|^2 = \left(\mathbf{a}-\mathbf{b}\right)^2 = \mathbf{a}^2 - 2\mathbf{a}\cdot \mathbf{b} + \mathbf{b}^2 \]
\[ \mathbf{a}^2 = 25\;\; \textit{(Definition of dot product)} \]
\[ \mathbf{b}^2 = 4\;\; \textit{(Definition of dot product)} \]
\[ \mathbf{a}\cdot \mathbf{b} = 5\cdot 2 \cdot \cos{\frac{\pi}{3}} = 5 \]
\[ |\mathbf{w}|^2 = 25 - 2\cdot 5 + 4 = 19 \]
\[ |\mathbf{w}| = \sqrt{19} \]
a)
\[ \mathbf{r}_B(10) = \left(\begin{array}{c} 0 \\ 5 \\ 0 \end{array}\right) + 10\left(\begin{array}{c} -2 \\ 6 \\ 3 \end{array}\right) = \left(\begin{array}{c} -20 \\ 65 \\ 30 \end{array}\right) \]
b) Their starting positions are \(\left(\begin{array}{c} 20 \\ 0 \\ 0 \end{array}\right)\) and \(\left(\begin{array}{c} 0 \\ 5 \\ 0 \end{array}\right)\) respectively. The distance between these two points is \(\sqrt{20^2 + 5^2 + 0^2} = 20.6\).
c) If they intersect, then \(\mathbf{r}_A = \mathbf{r}_B\), meaning the components of their position vectors should match.
\[ \begin{align*} 20 - 6t &= -2s \\ 0.5t &= 5 + 6s \\ t &= 3s \end{align*} \]
Solving the first two equations, we get \(t = \frac{22}{7}\) and \(s = -\frac{4}{7}\), which clearly do not satisfy the third equation, hence they do not intersect.
d) (i) The distance between the two submarines is:
\[ \mathbf{r}_A - \mathbf{r}_B = \left[ \left(\begin{array}{c} 20 \\ 0 \\ 0 \end{array}\right) + t\left(\begin{array}{c} -6 \\ 0.5 \\ 1 \end{array}\right)\right] - \left[\left(\begin{array}{c} 0 \\ 5 \\ 0 \end{array}\right) + t\left(\begin{array}{c} -2 \\ 6 \\ 3 \end{array}\right) \right] = \left(\begin{array}{c} 20 - 4t \\ -5 - 5.5t \\ -2t \end{array}\right) \]
The magnitude of this is: \[|\mathbf{r}_A - \mathbf{r}_B|(1.04) = \sqrt{(20 - 4t)^2 + (-5 - 5.5t)^2 + (-2t)^2}\]We can find the local minimum of this by either graphing it with our GDC or by taking the derivative and setting it equal to 0. Either way, we get \[t = 1.04\]
d) (ii) \[ |\mathbf{r}_A - \mathbf{r}_B|(1.04) = \sqrt{(20 - 4(1.04))^2 + (-5 - 5.5(1.04))^2 + (-2(1.04))^2} = 19.24 \]
a) The direction vectors of the two lines are \(\mathbf{r_1} = \left(\begin{array}{c} 2 \\ 1 \\ -3 \end{array}\right)\) and \(\mathbf{r_2} = \left(\begin{array}{c} -2 \\ 2 \\ -1 \end{array}\right)\). The angle between them can be found using the definition of the dot product:
\[ \cos{\theta} = \frac{-4 + 2 + 3}{\sqrt{4 + 1 + 9} \cdot \sqrt{4 + 4 + 1}} = -\frac{\sqrt{14}}{42} \]
\[\theta = 1.66 \text{ radians} = 95.11^{\circ}\]
b) When two lines intersect, their respective components take the same value:
\[ \begin{align*} 2\lambda &= -2 - 2\mu \\ 5 + \lambda &= 5 + 2\mu \end{align*} \]
Solving these simultaneously, we get \(\lambda = -\frac{2}{3}\) and \(\mu = -\frac{1}{3}\). Let's check whether they align with the \( z \) coordinate of the two lines as well, as only then will they have a common point:
\[ \begin{align*} 1 - 3\lambda &= \frac{8}{3} - \mu \\ 1 - 3\left(-\frac{2}{3}\right) &= \frac{8}{3} - \left(-\frac{1}{3}\right) \\ 3 &= 3 \end{align*} \]
c) The vector connecting the origin to \(\mathbf{L_2}\) is:
\[ OL_2 = \left(\begin{array}{c} -2 - 2\mu \\ 5 + 2\mu \\ \frac{8}{3} - \mu \end{array}\right) \]
When considering the closest point, the direction vector of \(\mathbf{OL_2}\) should be perpendicular to \(\mathbf{L_2}\), as that is when we have the closest point. Thus:
\[ \begin{align*} 0 &= \left(\begin{array}{c} -2 - 2\mu \\ 5 + 2\mu \\ \frac{8}{3} - \mu \end{array}\right) \cdot \left(\begin{array}{c} -2 \\ 2 \\ -1 \end{array}\right) \\ 0 &= \frac{34}{3} + 9\mu \\ \mu &= -\frac{34}{27} \end{align*} \]