Line Questions in English

(C) The lines are given by $\frac{x-4}{1} = \frac{y-3}{2} = \frac{z-2}{-3}$ and $\frac{x+2}{2} = \frac{y-6}{4} = \frac{z-5}{-5}$.
For line $1$,a point $P_1 = (4, 3, 2)$ and direction vector $\vec{v}_1 = (1, 2, -3)$.
For line $2$,a point $P_2 = (-2, 6, 5)$ and direction vector $\vec{v}_2 = (2, 4, -5)$.
The vector connecting the two points is $\vec{P_1P_2} = (-2-4, 6-3, 5-2) = (-6, 3, 3)$.
The cross product of the direction vectors is $\vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}(-10+12) - \hat{j}(-5+6) + \hat{k}(4-4) = (2, -1, 0)$.
The shortest distance $d$ is given by $d = \frac{|\vec{P_1P_2} \cdot (\vec{v}_1 \times \vec{v}_2)|}{|\vec{v}_1 \times \vec{v}_2|}$.
Calculating the dot product: $|(-6, 3, 3) \cdot (2, -1, 0)| = |-12 - 3 + 0| = |-15| = 15$.
Calculating the magnitude of the cross product: $|\vec{v}_1 \times \vec{v}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5}$.
Therefore,$d = \frac{15}{\sqrt{5}} = 3\sqrt{5}$.

(A) $1$. For point $P(0, -5, 0)$ and line $L_1: \frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2} = \lambda$,the foot of perpendicular $F_1$ is $(2\lambda+1, \lambda, -2\lambda-1)$. The vector $\vec{PF_1} = (2\lambda+1, \lambda+5, -2\lambda-1)$. Since $\vec{PF_1} \cdot (2, 1, -2) = 0$,we get $2(2\lambda+1) + 1(\lambda+5) - 2(-2\lambda-1) = 0$,which simplifies to $9\lambda + 9 = 0$,so $\lambda = -1$. Thus $F_1 = (-1, -1, 1)$. The image $R = 2F_1 - P = 2(-1, -1, 1) - (0, -5, 0) = (-2, 3, 2)$.
$2$. For point $Q(0, -1/2, 0)$ and line $L_2: \frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1} = \mu$,the foot of perpendicular $F_2$ is $(-\mu+1, 4\mu-9, \mu-1)$. The vector $\vec{QF_2} = (-\mu+1, 4\mu-8.5, \mu-1)$. Since $\vec{QF_2} \cdot (-1, 4, 1) = 0$,we get $1(\mu-1) + 16\mu - 34 + \mu - 1 = 0$,which simplifies to $18\mu - 36 = 0$,so $\mu = 2$. Thus $F_2 = (-1, -1, 1)$. The image $S = 2F_2 - Q = 2(-1, -1, 1) - (0, -0.5, 0) = (-2, -1.5, 2)$.
$3$. Vectors: $\vec{PQ} = (0, 4.5, 0)$ and $\vec{PS} = (-2, 3.5, 2)$.
$4$. Area of parallelogram $PQRS = |\vec{PQ} \times \vec{PS}| = |(0, 4.5, 0) \times (-2, 3.5, 2)| = |(9, 0, 9)| = \sqrt{81 + 0 + 81} = \sqrt{162}$.
$5$. Square of the area = $162$.