已知 $a$ 和 $b$ 为两个非零且不平行的向量，且 $c = |b|a + |a|b$，证明 $c$ 在 $a$ 上的投影长度与 $c$ 在 $b$ 上的投影长度大小相等。 直线 $l_1$，平面 $\pi_1$ 和 $\pi_2$ 的方程分别为 $l_1: \; \dfrac{x-3}{2} = \dfrac{1-y}{2} = -z$, $\pi_1: \; 2x + 3y - z = 5$, $\pi_2: \; -ax - 3y + 2z = b$。 在接下来的问题中，设 $l_1$ 不在 $\pi_2$ 上，且 $l_1$ 与 $\pi_1$ 交于点 F。 若 $l_1$ 在 $\pi_2$ 上，求 $a$ 和 $b$ 的值。 求 F 的坐标。 用 $a$ 表示出 $\pi_1$ 与 $\pi_2$ 的交线 $l_2$ 的一个方向向量。 若 F 也在 $\pi_2$ 上，求 $a$ 与 $b$ 之间的关系。用 $a$ 表示出 $l_2$ 的一个向量方程。

1. Calculate the scalar projection of $c$ onto $a$.

Let the scalar projection of $\mathbf{c}$ onto $\mathbf{a}$ be $p_a$.
$p_a = \mathbf{c} \cdot \frac{\mathbf{a}}{|\mathbf{a}|}$
$p_a = (|\mathbf{b}|\mathbf{a} + |\mathbf{a}|\mathbf{b}) \cdot \frac{\mathbf{a}}{|\mathbf{a}|}$
$p_a = \frac{1}{|\mathbf{a}|} \left( |\mathbf{b}|(\mathbf{a} \cdot \mathbf{a}) + |\mathbf{a}|(\mathbf{b} \cdot \mathbf{a}) \right)$
Since $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$ and $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$,
$p_a = \frac{1}{|\mathbf{a}|} \left( |\mathbf{b}||\mathbf{a}|^2 + |\mathbf{a}|(\mathbf{a} \cdot \mathbf{b}) \right)$
$p_a = |\mathbf{a}||\mathbf{b}| + \mathbf{a} \cdot \mathbf{b}$

2. Calculate the scalar projection of $c$ onto $b$.

Let the scalar projection of $\mathbf{c}$ onto $\mathbf{b}$ be $p_b$.
$p_b = \mathbf{c} \cdot \frac{\mathbf{b}}{|\mathbf{b}|}$
$p_b = (|\mathbf{b}|\mathbf{a} + |\mathbf{a}|\mathbf{b}) \cdot \frac{\mathbf{b}}{|\mathbf{b}|}$
$p_b = \frac{1}{|\mathbf{b}|} \left( |\mathbf{b}|(\mathbf{a} \cdot \mathbf{b}) + |\mathbf{a}|(\mathbf{b} \cdot \mathbf{b}) \right)$
Since $\mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2$,
$p_b = \frac{1}{|\mathbf{b}|} \left( |\mathbf{b}|(\mathbf{a} \cdot \mathbf{b}) + |\mathbf{a}||\mathbf{b}|^2 \right)$
$p_b = \mathbf{a} \cdot \mathbf{b} + |\mathbf{a}||\mathbf{b}|$

3. Compare the magnitudes of the two projections and conclude.

The length (magnitude) of the projection of $\mathbf{c}$ onto $\mathbf{a}$ is $|p_a| = ||\mathbf{a}||\mathbf{b}| + \mathbf{a} \cdot \mathbf{b}|$.

The length (magnitude) of the projection of $\mathbf{c}$ onto $\mathbf{b}$ is $|p_b| = |\mathbf{a} \cdot \mathbf{b} + |\mathbf{a}||\mathbf{b}||$.

Since $p_a = p_b$, their absolute values are exactly the same, meaning the lengths of the projections have the same magnitude. (Shown)

1. Find the direction vector of $l_1$ and the normal vector of $\pi_2$, and use their dot product to find $a$.

Rewriting the equation of $l_1$ in standard form: $\dfrac{x-3}{2} = \dfrac{y-1}{-2} = \dfrac{z-0}{-1}$.

The direction vector of $l_1$ is $\mathbf{d} = \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$.
The normal vector of $\pi_2$ is $\mathbf{n}_2 = \begin{pmatrix} -a \\ -3 \\ 2 \end{pmatrix}$.

If $l_1$ lies on $\pi_2$, it must be perpendicular to the normal of the plane, so $\mathbf{d} \cdot \mathbf{n}_2 = 0$:
$\begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix} \cdot \begin{pmatrix} -a \\ -3 \\ 2 \end{pmatrix} = 0$
$-2a + 6 - 2 = 0$
$2a = 4 \implies a = 2$

2. Use a point on $l_1$ to find $b$.

From the equation of $l_1$, a point on the line is $(3, 1, 0)$.
Since the line lies entirely on $\pi_2$, this point must satisfy the equation of $\pi_2$:
$-a(3) - 3(1) + 2(0) = b$
Substitute $a = 2$:
$-3(2) - 3 = b$
$b = -9$

1. Express the coordinates of any point on line $l_1$ in terms of a parameter.

Let $\frac{x-3}{2} = \frac{1-y}{2} = -z = \lambda$.
Then the parametric equations are:
$x = 3 + 2\lambda$
$y = 1 - 2\lambda$
$z = -\lambda$

2. Substitute the parametric coordinates into the equation of plane $\pi_1$ to find the value of $\lambda$ at the intersection point.

Substitute $x$, $y$, and $z$ into $2x + 3y - z = 5$:
$2(3 + 2\lambda) + 3(1 - 2\lambda) - (-\lambda) = 5$
$6 + 4\lambda + 3 - 6\lambda + \lambda = 5$
$9 - \lambda = 5$
$\lambda = 4$

3. Calculate the coordinates of F using the found value of $\lambda$.

Substitute $\lambda = 4$ into the parametric equations:
$x = 3 + 2(4) = 11$
$y = 1 - 2(4) = -7$
$z = -4$

The coordinates of F are $(11, -7, -4)$.

1. Identify the normal vectors of the planes $\pi_1$ and $\pi_2$.

From the Cartesian equations of the planes, the normal vectors are:
$\mathbf{n}_1 = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix}$ for $\pi_1$
$\mathbf{n}_2 = \begin{pmatrix} -a \\ -3 \\ 2 \end{pmatrix}$ for $\pi_2$

2. Calculate the cross product of the normal vectors to find the direction vector of the line of intersection.

The direction vector $\mathbf{d}$ of the line of intersection $l_2$ is parallel to $\mathbf{n}_1 \times \mathbf{n}_2$:
$\mathbf{d} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} \times \begin{pmatrix} -a \\ -3 \\ 2 \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} (3)(2) - (-1)(-3) \\ (-1)(-a) - (2)(2) \\ (2)(-3) - (3)(-a) \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} 6 - 3 \\ a - 4 \\ -6 + 3a \end{pmatrix}$
$\mathbf{d} = \begin{pmatrix} 3 \\ a - 4 \\ 3a - 6 \end{pmatrix}$

1. Find the coordinates of $F$ (from part ii) and substitute into $\pi_2$ to find the relationship between $a$ and $b$.

Expressing $l_1$ parametrically: $\mathbf{r} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$.
Substituting into $\pi_1$:
$2(3 + 2\lambda) + 3(1 - 2\lambda) - (-\lambda) = 5$
$6 + 4\lambda + 3 - 6\lambda + \lambda = 5 \implies \lambda = 4$.
Thus, $F$ is $(11, -7, -4)$.
Since $F$ also lies on $\pi_2: -ax - 3y + 2z = b$,
$-a(11) - 3(-7) + 2(-4) = b$
$-11a + 21 - 8 = b$
$11a + b = 13$

2. Determine the direction vector of $l_2$ (from part iii) and state the vector equation of $l_2$.

The direction vector $\mathbf{d}$ of $l_2$ is the cross product of the normal vectors of $\pi_1$ and $\pi_2$:
$\mathbf{d} = \begin{pmatrix} 2 \\ 3 \\ -1 \end{pmatrix} \times \begin{pmatrix} -a \\ -3 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 \\ a - 4 \\ 3a - 6 \end{pmatrix}$
Since $F$ lies on both $\pi_1$ and $\pi_2$, it must lie on their line of intersection $l_2$. Using $F$ as the position vector:
$\mathbf{r} = \begin{pmatrix} 11 \\ -7 \\ -4 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ a - 4 \\ 3a - 6 \end{pmatrix}, \mu \in \mathbb{R}$