The Cartesian equation of line $L_1$ is $\frac{x-2}{a}=\frac{y+2}{b}=\frac{z-3}{c}$, where $a, b, c$ are constants. The line $L_2$ is parallel to the vector $4 \mathbf{i}+3 \mathbf{j}$. The line $L_3$ passes through the origin and the point with position vector $\mathbf{j}+\mathbf{k}$.
(i) Given that $L_1$ is perpendicular to $L_2$, form an equation relating $a$ and $b$.
[1]
(ii) Given that $L_1$ intersects $L_3$, show that $5 a+2 b-2 c=0$.
[3]
(iii) Hence express $a$ and $b$ in terms of $c$.
[1]
(iv) Find the acute angle between $L_1$ and $L_3$.
[2]
(i) $4 a+3 b=0$
(iii) $b=-\frac{8}{7} c$
(iv) Acute angle between the two planes is $86.7^{\circ}$
(i)
$L_1: \frac{x-2}{a}=\frac{y+2}{b}=\frac{z-3}{c} $

$L_1$ is perpendicular to $L_2$,$

$4 a+3 b=0$
(ii)

$-2+\lambda b =\mu $
$3+\lambda c =\mu$
Sub (4) into (5):
$-2+\lambda b=3+\lambda c$
Sub (3) into (6):
$-2+\left(\frac{-2}{a}\right) b=3+\left(\frac{-2}{a}\right) c$
$-2 a-2 b=3 a-2 c$
$5 a+2 b-2 c=0 \quad \text { (Shown) }$
(iii)Using results in (i) \& (ii), use GC to solve
$4 a+3 b+0 c=0 $
$5 a+2 b-2 c=0 $
$a=\frac{6}{7} c $
$b=-\frac{8}{7} c$
(iv)
Using result in (iii),




$\theta=86.7^{\circ}$
Acute angle between the two planes is $86.7^{\circ}$