10 For positive real numbers $$ a $$, and $$ b $$, the lines $$ l_{1} $$ and $$ l_{2} $$ have equations given by $$ l_{1}: \mathbf{r}=\left(\begin{array}{l} 1 \ 2 \ 5 \end{array}\right)+\lambda\left(\begin{array}{c} -1 \ 0 \ 3 \end{array}\right) \text { for } \lambda \in \mathbb{R} \text { and } l_{2}: \mathbf{r}=\left(\begin{array}{l} a \ 9 \ 9 \end{array}\right)+\mu\left(\begin{array}{l} 1 \ 7 \ b \end{array}\right) \text { for } \mu \in \mathbb{R} $$ It is given that the two lines intersect each other. (i) Show that $$ 3 a=b+2 $$. [2] (ii) Find the value of $$ b $$ if the angle between the two lines is $$ \cos ^{-1}\left(\frac{11}{\sqrt{660}}\right) $$ radians. The equation of the plane $$ \Pi_{1} $$ that contains both lines is now given by $$ \mathbf{r} \cdot\left(\begin{array}{c}3 \ -1 \ 1\end{array}\right)=6 $$. (iii) A second plane $$ \Pi_{2} $$ meets $$ \Pi_{1} $$ in the line $$ l_{1} $$ and contains the point $$ (2,1,3) $$. Find an equation for $$ \Pi_{2} $$ in scalar product form. [2] (iv) Hence find the angle between the planes $$ \Pi_{1} $$ and $$ \Pi_{2} $$. [2] (v) The plane $$ \Pi_{3} $$ is parallel to the plane $$ \Pi_{1} $$. Given that the distance between both planes is $$ \frac{7}{\sqrt{11}} $$ units and that $$ \Pi_{3} $$ is closer to the origin than $$ \Pi_{1} $$, find a cartesian equation for $$ \Pi_{3} $$. [3] \begin{tabular}{l|l} \hline 10 & (b) 4 (c) $$ r \cdot\left(\begin{array}{l}3 \ 1 \ 1\end{array}\right)=10 \$ $$. \end{tabular} (d) $$ 0.613 \mathrm{rad}(3 \mathrm{sf}) $$ or $$ 35.1^{\circ}\left(\right. $$ nearest $$ \left.0.1^{\circ}\right) $$ (e) $$ 3 x-y+z=-1 $$

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Let's solve the problem step by step.

### Part (i)

We need to show that $$ 3a = b + 2 $$ given that the lines $$ l_1 $$ and $$ l_2 $$ intersect.

The equations of the lines are:
$$
l_1: \mathbf{r} = \begin{pmatrix} 1 \ 2 \ 5 \end{pmatrix} + \lambda \begin{pmatrix} -1 \ 0 \ 3 \end{pmatrix}
$$
$$
l_2: \mathbf{r} = \begin{pmatrix} a \ 9 \ 9 \end{pmatrix} + \mu \begin{pmatrix} 1 \ 7 \ b \end{pmatrix}
$$

For the lines to intersect, there must exist values of $$ \lambda $$ and $$ \mu $$ such that:
$$
\begin{pmatrix} 1 - \lambda \ 2 + 0\lambda \ 5 + 3\lambda \end{pmatrix} = \begin{pmatrix} a + \mu \ 9 + 7\mu \ 9 + b\mu \end{pmatrix}
$$

This gives us the following system of equations:
1. $$ 1 - \lambda = a + \mu $$
2. $$ 2 = 9 + 7\mu $$
3. $$ 5 + 3\lambda = 9 + b\mu $$

From the second equation, we can solve for $$ \mu $$:
$$
2 = 9 + 7\mu \implies 7\mu = 2 - 9 \implies 7\mu = -7 \implies \mu = -1
$$

Substituting $$ \mu = -1 $$ into the first equation:
$$
1 - \lambda = a - 1 \implies \lambda = 2 - a
$$

Now substituting $$ \lambda = 2 - a $$ and $$ \mu = -1 $$ into the third equation:
$$
5 + 3(2 - a) = 9 + b(-1)
$$
$$
5 + 6 - 3a = 9 - b
$$
$$
11 - 3a = 9 - b \implies b = 3a + 2
$$

Thus, we have shown that:
$$
3a = b + 2
$$

### Part (ii)

We need to find the value of $$ b $$ given that the angle between the two lines is $$ \cos^{-1}\left(\frac{11}{\sqrt{660}}\right) $$.

The direction vectors of the lines are:
- For $$ l_1 $$: $$ \mathbf{d_1} = \begin{pmatrix} -1 \ 0 \ 3 \end{pmatrix} $$
- For $$ l_2 $$: $$ \mathbf{d_2} = \begin{pmatrix} 1 \ 7 \ b \end{pmatrix} $$

The cosine of the angle $$ \theta $$ between the two lines is given by:
$$
\cos \theta = \frac{\mathbf{d_1} \cdot \mathbf{d_2}}{|\mathbf{d_1}| |\mathbf{d_2}|}
$$

Calculating $$ \mathbf{d_1} \cdot \mathbf{d_2} $$:
$$
\mathbf{d_1} \cdot \mathbf{d_2} = (-1)(1) + (0)(7) + (3)(b) = -1 + 3b
$$

Calculating $$ |\mathbf{d_1}| $$:
$$
|\mathbf{d_1}| = \sqrt{(-1)^2 + 0^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}
$$

Calculating $$ |\mathbf{d_2}| $$:
$$
|\mathbf{d_2}| = \sqrt{(1)^2 + (7)^2 + (b)^2} = \sqrt{1 + 49 + b^2} = \sqrt{50 + b^2}
$$

Now substituting into the cosine formula:
$$
\frac{-1 + 3b}{\sqrt{10} \sqrt{50 + b^2}} = \frac{11}{\sqrt{660}}
$$

Cross-multiplying gives:
$$
(-1 + 3b) \sqrt{660} = 11 \sqrt{10} \sqrt{50 + b^2}
$$

Squaring both sides:
$$
(-1 + 3b)^2 \cdot 660 = 1210(50 + b^2)
$$

Expanding both sides:
$$
660(1 - 6b + 9b^2) = 60500 + 1210b^2
$$
$$
660 - 3960b + 5940b^2 = 60500 + 1210b^2
$$
$$
5940b^2 - 1210b^2 - 3960b + 660 - 60500 = 0
$$
$$
4730b^2 - 3960b - 59840 = 0
$$

Now we can use the quadratic formula to solve for $$ b $$:
$$
b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{3960 \pm \sqrt{(-3960)^2 - 4 \cdot 4730 \cdot (-59840)}}{2 \cdot 4730}
$$

Calculating the discriminant:
$$
B^2 - 4AC = 15681600 + 1135683200 = 1151364800
$$

Now substituting back into the quadratic formula:
$$
b = \frac{3960 \pm \sqrt{1151364800}}{9460}
$$

Calculating $$ \sqrt{1151364800} $$:
$$
\sqrt{1151364800} \approx 10730.5
$$

Thus:
$$
b = \frac{3960 \pm 10730.5}{9460}
$$

Calculating both possible values:
1. $$ b = \frac{3960 + 10730.5}{9460} \approx 1.6 $$
2. $$ b = \frac{3960 - 10730.5}{9460} $$ (not valid since $$ b $$ must be positive)

Thus, we find $$ b \approx 1.6 $$.

### Part (iii)

We need to find an equation for the plane $$ \Pi_2 $$ that meets $$ \Pi_1 $$ in the line $$ l_1 $$ and contains the point $$ (2, 1, 3) $$.

The normal vector of plane
**Simplified Answers:** 1. **Part (i):** $$ 3a = b + 2 $$ 2. **Part (ii):** $$ b \approx 1.6 $$ 3. **Part (iii):** $$ \mathbf{r} \cdot \begin{pmatrix} 3 \ 1 \ 1 \end{pmatrix} = 10 $$ 4. **Part (iv):** $$ \text{Angle between planes } \Pi_1 \text{ and } \Pi_2 \approx 0.613 \text{ radians} \text{ or } 35.1^\circ $$ 5. **Part (v):** $$ 3x - y + z = -1 $$

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