Finding the vector equation of \(L_1\):

1. First, we calculate the direction vector \(\overrightarrow{AB}\) by subtracting coordinates of A from B:

\(\overrightarrow{AB} = \begin{pmatrix}2-1\\3-0\\-1-4\end{pmatrix} = \begin{pmatrix}1\\3\\-5\end{pmatrix}\)

2. Using point A and this direction vector, the vector equation is:

\(\boldsymbol{r} = \begin{pmatrix}1\\0\\4\end{pmatrix} + t\begin{pmatrix}1\\3\\-5\end{pmatrix}\)

[2 marks]

Showing \(L_1\) and \(L_2\) are skew lines:

1. Convert both lines to parametric form:

\(L_1\): \(x=1+t\), \(y=3t\), \(z=4-5t\)

\(L_2\): From symmetric form \(\frac{x-1}{3} = \frac{y+2}{1} = \frac{z-1}{-2} = s\), we get:

\(x=1+3s\), \(y=-2+s\), \(z=1-2s\)

2. Check for intersection by equating x and y components:

\(1+t=1+3s \Rightarrow t=3s\)

\(3t=-2+s\)

Substituting: \(3(3s)=-2+s \Rightarrow 9s=-2+s \Rightarrow s=-\frac{1}{4}\), \(t=-\frac{3}{4}\)

3. Verify z components with these parameters:

For \(L_1\): \(z=4-5(-\frac{3}{4})=\frac{31}{4}\)

For \(L_2\): \(z=1-2(-\frac{1}{4})=\frac{3}{2}\)

4. Since \(\frac{31}{4} \neq \frac{3}{2}\), the lines don’t intersect.

5. Check if direction vectors are parallel:

\(\begin{pmatrix}1\\3\\-5\end{pmatrix}\) and \(\begin{pmatrix}3\\1\\-2\end{pmatrix}\) are not scalar multiples, so lines are not parallel.

Conclusion: \(L_1\) and \(L_2\) are skew lines – they don’t intersect and aren’t parallel.

[5 marks]

Finding the equation of plane \(\Pi_1\):

1. Since \(\Pi_1\) is parallel to both \(L_1\) and \(L_2\), we use their direction vectors to find the normal vector:

\(\mathbf{n} = \begin{pmatrix}1\\3\\-5\end{pmatrix} \times \begin{pmatrix}3\\1\\-2\end{pmatrix}\)

Calculating cross product:

\(= \begin{pmatrix}(3)(-2)-(-5)(1)\\-[(1)(-2)-(-5)(3)]\\(1)(1)-(3)(3)\end{pmatrix} = \begin{pmatrix}-6+5\\-[-2+15]\\1-9\end{pmatrix} = \begin{pmatrix}-1\\-13\\-8\end{pmatrix}\)

2. Using point C(0,1,-2) in the plane:

Equation: \(-1(x-0)-13(y-1)-8(z+2)=0\)

Simplify: \(-x-13y+13-8z-16=0\)

Final equation: \(-x-13y-8z=3\) or equivalently \(x+13y+8z=-3\)

[4 marks]

(i) Finding \(k\):

1. The angle between a line and plane is the complement of the angle between the line and the normal vector.

Given angle between \(L_3\) and \(\Pi_2\) is 60°, the angle between \(L_3\) and normal \(\mathbf{n}=\begin{pmatrix}1\\1\\0\end{pmatrix}\) is 30°.

2. Using dot product formula:

\(\cos 30° = \frac{\begin{pmatrix}k\\1\\-1\end{pmatrix} \cdot \begin{pmatrix}1\\1\\0\end{pmatrix}}{|\begin{pmatrix}k\\1\\-1\end{pmatrix}||\begin{pmatrix}1\\1\\0\end{pmatrix}|}\)

\(\frac{\sqrt{3}}{2} = \frac{k+1}{\sqrt{k^2+1+1}\sqrt{1+1+0}} = \frac{k+1}{\sqrt{2(k^2+2)}}\)

3. Square both sides:

\(\frac{3}{4} = \frac{(k+1)^2}{2(k^2+2)}\)

Cross-multiply: \(3(k^2+2) = 2(k^2+2k+1)\)

Simplify: \(3k^2+6 = 2k^2+4k+2\)

Final equation: \(k^2-4k+4=0\)

Solution: \(k=2\) (double root)

(ii) Finding intersection point P:

1. With \(k=2\), \(L_3\) becomes: \(\boldsymbol{r} = \begin{pmatrix}3\\0\\1\end{pmatrix} + \lambda \begin{pmatrix}2\\1\\-1\end{pmatrix}\)

2. Substitute into \(\Pi_2\): \(x + y = 12\)

\((3+2\lambda) + (0+\lambda) = 12\)

Simplify: \(3 + 3\lambda = 12 \Rightarrow \lambda = 3\)

3. Find coordinates:

\(x=3+2(3)=9\), \(y=0+3=3\), \(z=1-3=-2\)

Thus, \(P(9,3,-2)\)

[7 marks]

1. Vector Geometry: Understanding how to derive line equations from points and direction vectors is fundamental.
2. Skew Lines: Two lines in 3D space are skew if they are neither parallel nor intersecting.
3. Plane Equations: A plane can be defined using a normal vector and a point, with the normal found via cross product of two direction vectors.
4. Angles in 3D: The angle between a line and plane is complementary to the angle between the line and the plane’s normal vector.