Detailed Answer

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:

\[ \cos(\theta) = \frac{825}{\sqrt{15^2 + 60^2} \cdot \sqrt{35^2 + 5^2}} = 67.84^\circ \]

Detailed Answer

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:

\[|\mathbf{PC}| = \sqrt{\left(-200 + 35(4.03)\right)^2 + \left(-110 + 40(4.03)\right)^2} = 78.1 \, \text{km} > 50 \, \text{km}\]

Detailed Answer

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.

Detailed Answer

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)

\[ \mathbf{F} = a\left(\begin{array}{c} 17 \\ -2 \\ -6 \end{array}\right) \times \left(\begin{array}{c} 2 \\ 5 \\ 4 \end{array}\right) = a\left(\begin{array}{c} 22 \\ -80 \\ 89 \end{array}\right) \]

\[ |\mathbf{F}| = a\sqrt{22^2 + 80^2 + 89^2} \approx 122a = 61 \]

\[ a = 2 \]

Detailed Answer

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 \(\mathbf{r}\).

\[ \mathbf{r} = \left(\begin{array}{c} \frac{74}{89} \\ -\frac{484}{89} \\ \frac{215}{89} \end{array}\right) \]

Detailed Answer

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:

\[ \text{Area} = \frac{1}{2} \left|\overrightarrow{\text{XY}} \times \overrightarrow{\text{XZ}}\right| = \frac{1}{2} \sqrt{9 + 4 + 1} = \frac{\sqrt{14}}{2} \]

Detailed Answer

\[ |\mathbf{w}|^2 = \left(\mathbf{a}-\mathbf{b}\right)^2 = \mathbf{a}^2 - 2\mathbf{a}\cdot \mathbf{b} + \mathbf{b}^2 \]

\[ \mathbf{a}^2 = 25\;\; \textit{(Definition of dot product)} \]

\[ \mathbf{b}^2 = 4\;\; \textit{(Definition of dot product)} \]

\[ \mathbf{a}\cdot \mathbf{b} = 5\cdot 2 \cdot \cos{\frac{\pi}{3}} = 5 \]

\[ |\mathbf{w}|^2 = 25 - 2\cdot 5 + 4 = 19 \]

\[ |\mathbf{w}| = \sqrt{19} \]

Detailed Answer

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) \[ |\mathbf{r}_A - \mathbf{r}_B|(1.04) = \sqrt{(20 - 4(1.04))^2 + (-5 - 5.5(1.04))^2 + (-2(1.04))^2} = 19.24 \]

Detailed Answer

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:

\[ \begin{align*} 0 &= \left(\begin{array}{c} -2 - 2\mu \\ 5 + 2\mu \\ \frac{8}{3} - \mu \end{array}\right) \cdot \left(\begin{array}{c} -2 \\ 2 \\ -1 \end{array}\right) \\ 0 &= \frac{34}{3} + 9\mu \\ \mu &= -\frac{34}{27} \end{align*} \]

Detailed Answer

a) Given: \[ \mathbf{u} = 2\mathbf{i} - 3\mathbf{j}, \quad \mathbf{v} = -\mathbf{i} + 4\mathbf{j}. \]

Calculate \[ \mathbf{u} + 3\mathbf{v} = (2\mathbf{i} - 3\mathbf{j}) + 3(-\mathbf{i} + 4\mathbf{j}). \]

Expand and simplify: \[ \mathbf{u} + 3\mathbf{v} = 2\mathbf{i} - 3\mathbf{j} - 3\mathbf{i} + 12\mathbf{j}. \]

Combine like terms: \[ \mathbf{u} + 3\mathbf{v} = (2 - 3)\mathbf{i} + (-3 + 12)\mathbf{j} = -\mathbf{i} + 9\mathbf{j}. \]

b) The vector \[ \mathbf{w} = \frac{-25\sqrt{82}}{82}\mathbf{i} + \frac{225\sqrt{82}}{82}\mathbf{j} \]

Detailed Answer

a) The scalar product of two vectors \[ \begin{pmatrix} a \\ b \end{pmatrix} \cdot \begin{pmatrix} c \\ d \end{pmatrix} = a \cdot c + b \cdot d. \]

For the vectors \[ \begin{pmatrix} 60 \\ 25 \end{pmatrix} \text{ and } \begin{pmatrix} -30 \\ 40 \end{pmatrix}, \] we have: \[ \begin{pmatrix} 60 \\ 25 \end{pmatrix} \cdot \begin{pmatrix} -30 \\ 40 \end{pmatrix} = 60 \cdot (-30) + 25 \cdot 40 = -1800 + 1000 = -800. \]

b) (i) The distance between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \) is given by: \[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. \]

For \( P(60, 25) \) and \( Q(-30, 40) \), we have: \[ \text{Distance} = \sqrt{(-30 - 60)^2 + (40 - 25)^2} = \sqrt{(-90)^2 + (15)^2} = \sqrt{8100 + 225} = \sqrt{8325}. \]

Simplifying \( \sqrt{8325} \): \[ \sqrt{8325} = \sqrt{25 \times 333} = 5\sqrt{333}. \]

b) (ii) The angle \( \theta \) between the vectors \( \vec{OP} \) and \( \vec{OQ} \) can be found using the scalar product formula: \[ \cos \theta = \frac{\vec{OP} \cdot \vec{OQ}}{|\vec{OP}| \cdot |\vec{OQ}|} \]

First, compute the magnitudes of \( \vec{OP} \) and \( \vec{OQ} \): \[ |\vec{OP}| = \sqrt{60^2 + 25^2} = \sqrt{3600 + 625} = \sqrt{4225} = 65, \] and \[ |\vec{OQ}| = \sqrt{(-30)^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \]

From part (a), the scalar product is \( \vec{OP} \cdot \vec{OQ} = -800 \). Thus: \[ \cos \theta = \frac{-800}{65 \cdot 50} = \frac{-800}{3250} = -\frac{16}{65} \]

Now, find \( \theta \): \[ \theta = \cos^{-1}\left(-\frac{16}{65}\right) \]

Using a calculator (GDC), we find: \[ \theta \approx 104.5^\circ \]

Detailed Answer

a) (i) The velocity vector \( \vec{AB} \) is the displacement vector from \( A \) to \( B \). The displacement vector is: \[ \vec{AB} = \begin{pmatrix} 12 - 8 \\ -7 - (-3) \\ 18 - 12 \end{pmatrix} = \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix} \] Thus, the velocity vector is: \[ \vec{v} = \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix} \]

a) (ii) The speed of the car is the magnitude of the velocity vector \( \vec{v} \): \[ \text{Speed} = |\vec{v}| = \sqrt{4^2 + (-4)^2 + 6^2} = \sqrt{16 + 16 + 36} = \sqrt{68} \] Simplifying \( \sqrt{68} \): \[ \sqrt{68} = \sqrt{4 \cdot 17} = 2\sqrt{17} \]

b) The equation of motion of the car can be written in vector form as: \[ \vec{r}(t) = \vec{r}_0 + t \cdot \vec{v} \] where:

• \( \vec{r}(t) \) is the position of the car at time \( t \),
• \( \vec{r}_0 \) is the initial position of the car,
• \( \vec{v} \) is the velocity vector.

Substituting the known values:

\[ \vec{r}(t) = \begin{pmatrix} 8 \\ -3 \\ 12 \end{pmatrix} + t \cdot \begin{pmatrix} 4 \\ -4 \\ 6 \end{pmatrix} \]

Thus, the equation of motion is:

\[ \vec{r}(t) = \begin{pmatrix} 8 + 4t \\ -3 - 4t \\ 12 + 6t \end{pmatrix} \]

Detailed Answer

a) The velocity vector of the first drone is \( \begin{pmatrix} 2 \\ 5 \\ 6 \end{pmatrix} \) km/h. The speed is the magnitude of this vector: \[ \text{Speed} = \sqrt{2^2 + 5^2 + 6^2} = \sqrt{4 + 25 + 36} = \sqrt{65} \] Thus, the speed of the first drone is \( \sqrt{65} \) km/h.

b) The second drone moves from \( (-3, 12, 20) \) to \( (0, 18, 29) \) in 1.5 hours. The displacement vector is: \[ \begin{pmatrix} 0 - (-3) \\ 18 - 12 \\ 29 - 20 \end{pmatrix} = \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} \] The velocity vector is the displacement divided by the time (1.5 hours): \[ \begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix} \div 1.5 = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} \] Thus, the position vector after \( t \) hours is: \[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \\ 20 \end{pmatrix} + t \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} \]

c) (i) Set the position vectors of the two drones equal to each other: \[ \begin{pmatrix} 5 \\ -2 \\ 8 \end{pmatrix} + t \begin{pmatrix} 2 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -3 \\ 12 \\ 20 \end{pmatrix} + t \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} \] This gives the system of equations: \[ \begin{cases} 5 + 2t = -3 + 2t, \\ -2 + 5t = 12 + 4t, \\ 8 + 6t = 20 + 6t \end{cases} \] From the first equation: \[ 5 + 2t = -3 + 2t \implies 5 = -3 \] which is impossible. This suggests that the drones do not meet.

c) (ii) Since the drones do not meet, there is no point \( Q \).

d) The angle \( \theta \) between the paths of the two drones is the angle between their velocity vectors \( \vec{v}_1 = \begin{pmatrix} 2 \\ 5 \\ 6 \end{pmatrix} \) and \( \vec{v}_2 = \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} \). The scalar product formula is: \[ \cos \theta = \frac{\vec{v}_1 \cdot \vec{v}_2}{|\vec{v}_1| \cdot |\vec{v}_2|} \] Compute the scalar product: \[ \vec{v}_1 \cdot \vec{v}_2 = 2 \cdot 2 + 5 \cdot 4 + 6 \cdot 6 = 4 + 20 + 36 = 60 \] Compute the magnitudes: \[ |\vec{v}_1| = \sqrt{2^2 + 5^2 + 6^2} = \sqrt{4 + 25 + 36} = \sqrt{65} \] \[ |\vec{v}_2| = \sqrt{2^2 + 4^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56} \] Thus: \[ \cos \theta = \frac{60}{\sqrt{65} \cdot \sqrt{56}} = \frac{60}{\sqrt{3640}} \] Now, find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{30}{\sqrt{910}}\right) \] Using a calculator (GDC), we find: \[ \theta \approx 14.5^\circ \]

Detailed Answer

a) The vector \( \overrightarrow{PQ} \) is: \[ \overrightarrow{PQ} = Q - P = (1 - 2, 0 - 3, 5 - 4) = (-1, -3, 1)\] The vector \( \overrightarrow{QR} \) is: \[ \overrightarrow{QR} = R - Q = (3 - 1, 1 - 0, 6 - 5) = (2, 1, 1) \]

b) The cross product is: \[ \overrightarrow{PQ} \times \overrightarrow{QR} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -3 & 1 \\ 2 & 1 & 1 \end{vmatrix} \] Expand the determinant: \[ \overrightarrow{PQ} \times \overrightarrow{QR} = \mathbf{i}((-3)(1) - (1)(1)) - \mathbf{j}((-1)(1) - (1)(2)) + \mathbf{k}((-1)(1) - (-3)(2)) \] Simplify: \[ \overrightarrow{PQ} \times \overrightarrow{QR} = \mathbf{i}(-3 - 1) - \mathbf{j}(-1 - 2) + \mathbf{k}(-1 + 6) = -4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k} \] Thus: \[ \overrightarrow{PQ} \times \overrightarrow{QR} = \begin{pmatrix} -4 \\ 3 \\ 5 \end{pmatrix} \]

c) The area of triangle \( PQR \) is half the magnitude of the cross product: \[ \text{Area} = \frac{1}{2} \left| \overrightarrow{PQ} \times \overrightarrow{QR} \right| = \frac{1}{2} \sqrt{(-4)^2 + 3^2 + 5^2} = \frac{1}{2} \sqrt{16 + 9 + 25} = \frac{1}{2} \sqrt{50} \] Simplify: \[ \text{Area} = \frac{1}{2} \cdot 5\sqrt{2} = \frac{5\sqrt{2}}{2} \]

d) The line \( L \) passes through point \( S(4, -2, -7) \) and is perpendicular to the plane containing triangle \( PQR \). The direction vector of \( L \) is the normal vector to the plane, which is \( \overrightarrow{PQ} \times \overrightarrow{QR} = \begin{pmatrix} -4 \\ 3 \\ 5 \end{pmatrix} \). The parametric equations of \( L \) are: \[ \begin{cases} x = 4 - 4t, \\ y = -2 + 3t, \\ z = -7 + 5t, \end{cases} \] where \( t \) is a scalar parameter.

Detailed Answer

The line \( L \) has the direction vector \( \vec{d} = \begin{pmatrix} 4 \\ -2 \\ 1 \end{pmatrix} \). The plane contains the point \( B(2, -3, 5) \) and the line \( L \). To find a second direction vector in the plane, we can use a point on the line \( L \). Let \( s = 0 \), so a point on \( L \) is \( C(3, 5, 8) \). The vector \( \overrightarrow{BC} \) is: \[ \overrightarrow{BC} = C - B = (3 - 2, 5 - (-3), 8 - 5) = (1, 8, 3) \]

The normal vector \( \vec{n} \) to the plane is the cross product of \( \vec{d} \) and \( \overrightarrow{BC} \): \[ \vec{n} = \vec{d} \times \overrightarrow{BC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -2 & 1 \\ 1 & 8 & 3 \end{vmatrix} \] Expand the determinant: \[ \vec{n} = \mathbf{i}((-2)(3) - (1)(8)) - \mathbf{j}((4)(3) - (1)(1)) + \mathbf{k}((4)(8) - (-2)(1)) \] Simplify: \[ \vec{n} = \mathbf{i}(-6 - 8) - \mathbf{j}(12 - 1) + \mathbf{k}(32 + 2) = -14\mathbf{i} - 11\mathbf{j} + 34\mathbf{k} \] Thus, the normal vector is: \[ \vec{n} = \begin{pmatrix} -14 \\ -11 \\ 34 \end{pmatrix} \]

The Cartesian equation of the plane is given by: \[ \vec{n} \cdot (\vec{r} - \vec{B}) = 0 \] where \( \vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \) and \( \vec{B} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} \). Substitute the known values: \[ \begin{pmatrix} -14 \\ -11 \\ 34 \end{pmatrix} \cdot \begin{pmatrix} x - 2 \\ y + 3 \\ z - 5 \end{pmatrix} = 0 \] Compute the dot product: \[ -14(x - 2) - 11(y + 3) + 34(z - 5) = 0 \] Expand and simplify: \[ -14x + 28 - 11y - 33 + 34z - 170 = 0 \implies -14x - 11y + 34z - 175 = 0 \] Multiply through by \(-1\) to make the equation look better (not necessary): \[ 14x + 11y - 34z + 175 = 0 \]

Detailed Answer

a) The plane \( \pi_1 \) has a normal vector \( \vec{n} = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} \) and passes through the point \( Q(2, 3, 10) \). The Cartesian equation of \( \pi_1 \) is given by: \[ \vec{n} \cdot (\vec{r} - \vec{Q}) = 0 \] where \( \vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \). Substituting the known values: \[ \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} x - 2 \\ y - 3 \\ z - 10 \end{pmatrix} = 0 \] Compute the dot product: \[ 4(x - 2) - 5(y - 3) + 2(z - 10) = 0 \] Expand and simplify: \[ 4x - 8 - 5y + 15 + 2z - 20 = 0 \implies 4x - 5y + 2z - 13 = 0 \] Thus, the Cartesian equation of \( \pi_1 \) is: \[ 4x - 5y + 2z - 13 = 0 \]

b) (i) The plane \( \pi_2 \) has the equation \( 2x + 4y - z = 5 \). Substitute the coordinates of \( Q(2, 3, 10) \) into the equation: \[ 2(2) + 4(3) - 10 = 4 + 12 - 10 = 6 \neq 5 \] This shows that \( Q \) does not lie in \( \pi_2 \).
b) (ii) The line of intersection of \( \pi_1 \) and \( \pi_2 \) has a direction vector \( \vec{d} \) that is perpendicular to both normal vectors \( \vec{n}_1 = \begin{pmatrix} 4 \\ -5 \\ 2 \end{pmatrix} \) and \( \vec{n}_2 = \begin{pmatrix} 2 \\ 4 \\ -1 \end{pmatrix} \). Thus: \[ \vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -5 & 2 \\ 2 & 4 & -1 \end{vmatrix} \] Expand the determinant: \[ \vec{d} = \mathbf{i}((-5)(-1) - (2)(4)) - \mathbf{j}((4)(-1) - (2)(2)) + \mathbf{k}((4)(4) - (-5)(2)) \] Simplify: \[ \vec{d} = \mathbf{i}(5 - 8) - \mathbf{j}(-4 - 4) + \mathbf{k}(16 + 10) = -3\mathbf{i} + 8\mathbf{j} + 26\mathbf{k} \] Thus, the direction vector is: \[ \vec{d} = \begin{pmatrix} -3 \\ 8 \\ 26 \end{pmatrix} \] To find a point on the line of intersection, solve the system of equations: \[ \begin{cases} 4x - 5y + 2z = 13, \\ 2x + 4y - z = 7 \end{cases} \] Let \( x = 0 \). Then: \[ \begin{cases} -5y + 2z = 13, \\ 4y - z = 7 \end{cases} \] Solve the second equation for \( z \): \[ z = 4y - 7 \] Substitute into the first equation: \[ -5y + 2(4y - 7) = 13 \implies -5y + 8y - 14 = 13 \implies 3y = 27 \implies y = 9 \] Thus, \( z = 4(9) - 7 = 29 \). A point on the line is \( (0, 9, 29) \). The vector equation of the line is: \[ \vec{r} = \begin{pmatrix} 0 \\ 9 \\ 29 \end{pmatrix} + t \begin{pmatrix} -3 \\ 8 \\ 26 \end{pmatrix} \]

c) The acute angle \( \theta \) between the planes is given by: \[ \cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| \cdot |\vec{n}_2|} \] Compute the dot product: \[ \vec{n}_1 \cdot \vec{n}_2 = 4(2) + (-5)(4) + 2(-1) = 8 - 20 - 2 = -14 \] Compute the magnitudes: \[ |\vec{n}_1| = \sqrt{4^2 + (-5)^2 + 2^2} = \sqrt{16 + 25 + 4} = \sqrt{45}, \] \[ |\vec{n}_2| = \sqrt{2^2 + 4^2 + (-1)^2} = \sqrt{4 + 16 + 1} = \sqrt{21} \] Thus: \[ \cos \theta = \frac{14}{\sqrt{45} \cdot \sqrt{21}} = \frac{14}{\sqrt{945}}\] Simplify \( \sqrt{945} \): \[ \sqrt{945} = \sqrt{9 \times 105} = 3\sqrt{105} \] Thus: \[ \cos \theta = \frac{14}{3\sqrt{105}} \] Now, find \( \theta \): \[ \theta = \cos^{-1}\left(\frac{14}{3\sqrt{105}}\right) \] Using a calculator (GDC), we find: \[ \theta \approx 45.2^\circ \]