1. **Problem statement:** Given the family of lines $g_a:\ \vec{x} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix} + r \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}, \ r,a \in \mathbb{R}$, analyze various properties of these lines. 2. **General form and rules:** Each line $g_a$ is defined by a point $\vec{p} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix}$ and a direction vector $\vec{d}_a = \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}$. The parameter $r$ moves along the line. --- **a) Describe the position of the lines and sketch for $a=0,1,2$:** - For $a=0$, $\vec{d}_0 = \begin{pmatrix}0 \\ 2 \\ 4\end{pmatrix}$. - For $a=1$, $\vec{d}_1 = \begin{pmatrix}1 \\ 2 \\ 2\end{pmatrix}$. - For $a=2$, $\vec{d}_2 = \begin{pmatrix}2 \\ 2 \\ 0\end{pmatrix}$. These lines all pass through the fixed point $(5,1,4)$ but have different directions depending on $a$. The direction vector changes linearly with $a$. --- **b) Which line is parallel to $\vec{v} = \begin{pmatrix}3 \\ 1 \\ -4\end{pmatrix}$?** Lines are parallel if their direction vectors are scalar multiples: $$\vec{d}_a = t \vec{v} \implies \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix} = t \begin{pmatrix}3 \\ 1 \\ -4\end{pmatrix}$$ From components: 1. $a = 3t$ 2. $2 = t$ 3. $4 - 2a = -4t$ From (2), $t=2$. Substitute into (1): $a=3 \times 2=6$. Check (3): $4 - 2 \times 6 = 4 - 12 = -8$ and $-4 \times 2 = -8$, consistent. **Answer:** The line with $a=6$ is parallel to $\vec{v}$. --- **c) Which line passes through $P(x,-3,1)$? Find $x$.** The line $g_a$ passes through $P$ if there exists $r$ such that: $$\begin{pmatrix}x \\ -3 \\ 1\end{pmatrix} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix} + r \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}$$ Equate components: 1. $x = 5 + r a$ 2. $-3 = 1 + 2r \implies 2r = -4 \implies r = -2$ 3. $1 = 4 + r(4 - 2a) \implies 1 - 4 = r(4 - 2a) \implies -3 = -2(4 - 2a)$ Simplify (3): $$-3 = -2(4 - 2a) = -8 + 4a \implies 4a = 5 \implies a = \frac{5}{4}$$ Now find $x$ from (1): $$x = 5 + (-2) \times \frac{5}{4} = 5 - \frac{10}{4} = 5 - 2.5 = 2.5$$ **Answer:** $x = 2.5$ and $a = \frac{5}{4}$. --- **d) Which line intersects the z-axis? Find the intersection point.** The z-axis has $x=0$, $y=0$. Find $r,a$ such that: $$\begin{pmatrix}0 \\ 0 \\ z\end{pmatrix} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix} + r \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}$$ From $x$ and $y$: 1. $0 = 5 + r a \implies r a = -5$ 2. $0 = 1 + 2r \implies 2r = -1 \implies r = -\frac{1}{2}$ Substitute $r$ into (1): $$-\frac{1}{2} a = -5 \implies a = 10$$ Find $z$: $$z = 4 + r(4 - 2a) = 4 - \frac{1}{2} (4 - 20) = 4 - \frac{1}{2}(-16) = 4 + 8 = 12$$ **Answer:** The line with $a=10$ intersects the z-axis at $(0,0,12)$. --- **e) Which line intersects the y-axis? Find the intersection point.** The y-axis has $x=0$, $z=0$. Find $r,a$ such that: $$\begin{pmatrix}0 \\ y \\ 0\end{pmatrix} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix} + r \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}$$ From $x$ and $z$: 1. $0 = 5 + r a \implies r a = -5$ 2. $0 = 4 + r(4 - 2a) \implies r(4 - 2a) = -4$ From (1), $r = -\frac{5}{a}$ (assuming $a \neq 0$). Substitute into (2): $$-\frac{5}{a} (4 - 2a) = -4 \implies \frac{5}{a} (4 - 2a) = 4$$ Multiply both sides by $a$: $$5(4 - 2a) = 4a$$ $$20 - 10a = 4a$$ $$20 = 14a \implies a = \frac{10}{7}$$ Then $r = -\frac{5}{a} = -\frac{5}{10/7} = -\frac{5 \times 7}{10} = -\frac{35}{10} = -3.5$ Find $y$: $$y = 1 + 2r = 1 + 2(-3.5) = 1 - 7 = -6$$ **Answer:** The line with $a=\frac{10}{7}$ intersects the y-axis at $(0,-6,0)$. --- **f) Which line intersects the x-axis? Find the intersection point.** The x-axis has $y=0$, $z=0$. Find $r,a$ such that: $$\begin{pmatrix}x \\ 0 \\ 0\end{pmatrix} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix} + r \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}$$ From $y$ and $z$: 1. $0 = 1 + 2r \implies r = -\frac{1}{2}$ 2. $0 = 4 + r(4 - 2a) \implies r(4 - 2a) = -4$ Substitute $r = -\frac{1}{2}$ into (2): $$-\frac{1}{2} (4 - 2a) = -4 \implies 4 - 2a = 8 \implies -2a = 4 \implies a = -2$$ Find $x$: $$x = 5 + r a = 5 + (-\frac{1}{2})(-2) = 5 + 1 = 6$$ **Answer:** The line with $a=-2$ intersects the x-axis at $(6,0,0)$. --- **g) Which lines are parallel to a coordinate plane?** A line is parallel to a coordinate plane if its direction vector is orthogonal to the normal vector of that plane. - $xy$-plane normal: $\vec{n} = \begin{pmatrix}0 \\ 0 \\ 1\end{pmatrix}$ - $yz$-plane normal: $\vec{n} = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$ - $xz$-plane normal: $\vec{n} = \begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$ Check dot products: 1. Parallel to $xy$-plane if $\vec{d}_a \cdot \vec{n} = 0$: $$a \cdot 0 + 2 \cdot 0 + (4 - 2a) \cdot 1 = 4 - 2a = 0 \implies a = 2$$ 2. Parallel to $yz$-plane if $\vec{d}_a \cdot \vec{n} = 0$: $$a \cdot 1 + 2 \cdot 0 + (4 - 2a) \cdot 0 = a = 0$$ 3. Parallel to $xz$-plane if $\vec{d}_a \cdot \vec{n} = 0$: $$a \cdot 0 + 2 \cdot 1 + (4 - 2a) \cdot 0 = 2 \neq 0$$ No $a$ satisfies this. **Answer:** Lines with $a=2$ are parallel to the $xy$-plane, lines with $a=0$ are parallel to the $yz$-plane, no line is parallel to the $xz$-plane. --- **h) The line $g_a$ contains point $P_a = (x_a,y_a,0)$ with $a \neq 2$. Find $x_a,y_a$ in terms of $a$ and show all $P_a$ lie on a line. Find that line's equation.** Since $P_a$ lies on $g_a$, there exists $r$ such that: $$\begin{pmatrix}x_a \\ y_a \\ 0\end{pmatrix} = \begin{pmatrix}5 \\ 1 \\ 4\end{pmatrix} + r \begin{pmatrix}a \\ 2 \\ 4 - 2a\end{pmatrix}$$ From $z$-component: $$0 = 4 + r(4 - 2a) \implies r(4 - 2a) = -4 \implies r = \frac{-4}{4 - 2a}$$ Since $a \neq 2$, denominator $\neq 0$. Find $x_a$ and $y_a$: $$x_a = 5 + r a = 5 + a \cdot \frac{-4}{4 - 2a} = 5 - \frac{4a}{4 - 2a}$$ $$y_a = 1 + 2r = 1 + 2 \cdot \frac{-4}{4 - 2a} = 1 - \frac{8}{4 - 2a}$$ Rewrite denominator: $$4 - 2a = 2(2 - a)$$ So: $$x_a = 5 - \frac{4a}{2(2 - a)} = 5 - \frac{2a}{2 - a}$$ $$y_a = 1 - \frac{8}{2(2 - a)} = 1 - \frac{4}{2 - a}$$ Set $t = 2 - a$, then: $$x_a = 5 - \frac{2(2 - t)}{t} = 5 - \frac{4 - 2t}{t} = 5 - \frac{4}{t} + 2 = 7 - \frac{4}{t}$$ $$y_a = 1 - \frac{4}{t}$$ So: $$x_a = 7 - \frac{4}{t}, \quad y_a = 1 - \frac{4}{t}$$ Eliminate $\frac{4}{t}$: $$\frac{4}{t} = 7 - x_a = 1 - y_a$$ Set equal: $$7 - x_a = 1 - y_a \implies y_a = x_a - 6$$ Also, $z_a = 0$. **Conclusion:** All points $P_a$ lie on the line in the plane $z=0$ with equation: $$y = x - 6, \quad z=0$$ This is a line in the $xy$-plane. --- **Final answers:** - b) $a=6$ - c) $a=\frac{5}{4}$, $x=2.5$ - d) $a=10$, intersection $(0,0,12)$ - e) $a=\frac{10}{7}$, intersection $(0,-6,0)$ - f) $a=-2$, intersection $(6,0,0)$ - g) $a=2$ parallel to $xy$-plane, $a=0$ parallel to $yz$-plane - h) $P_a$ lie on $y = x - 6$, $z=0$ line