Dr. Xiaozhou Li

Dr. Xiaozhou Li

Mathematics. Strength. Long-Term Mindset. Think Hard. Train Harder. Live Long.

Problem Sets: Vectors and the Geometry of Space (空间向量与解析几何)

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Last Update: 2026-04-05

  • $\vec{a} + 3\vec{b}$ and $\vec{a} - 4\vec{b}$ are perpendicular to vectors $7\vec{a} - 5\vec{b}$ and $7\vec{a} - 2\vec{b}$, respectively. Find the angle between $\vec{a}$ and $\vec{b}$.

  • Let $\vec{a} = (1, -1, 1)$, $\vec{b} = (3, -4, 5)$, and $\vec{x} = \vec{a} + \lambda\vec{b}$ where $\lambda$ is a real number. Prove that the vector $\vec{x}$ that minimizes $\lvert \vec{x}\rvert$ must be perpendicular to $\vec{b}$.

  • Let $\vec{a}$, $\vec{b}$, $\vec{c}$ be non-zero vectors, pairwise non-collinear, but $\vec{a} + \vec{b}$ is collinear with $\vec{c}$, and $\vec{b} + \vec{c}$ is collinear with $\vec{a}$. Prove that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

  • Find a line $L$ containsd in the plane $x + y + z = 1$ that is orthogonal to and intersects the line $L_0 = \begin{cases} y = 1, \ z = -1 \end{cases}$.

  • Find the line $L$ that is parallel to the planes $2x + 3y - 5 = 0$ and $y + z = 0$, and intersects the lines $\displaystyle L_1: \frac{x - 6}{3} = \frac{y}{2} = z - 1$ and $\displaystyle L_2: \frac{x}{3} = \frac{y - 8}{2} = \frac{z + 4}{-2}$.

  • Find the equation of the line passing through the point $P(-3, 5, -9)$ and intersecting the two lines $L_1: \begin{cases} y = 3x + 5, \ z = 2x - 3 \end{cases}$, $L_2: \begin{cases} y = 4x - 7, \ z = 5x + 10 \end{cases}$.

  • Find the equation of the plane passing through the point $P(2, 1, -1)$ and perpendicular to the line $\begin{cases} 2x + y - z = 3, \ x + 2y + z = 2 \end{cases}$.

  • Find the symmetric point of the origin with respect to the plane $6x + 2y - 9z + 121 = 0$.

  • Find the equation of the plane that passes through the line of intersection of the planes $x + 5y + z = 0$ and $x - z + 4 = 0$, and makes a $45^\circ$ angle with the plane $x - 4y - 8z + 12 = 0$.

  • Find the distance between the two skew lines $\displaystyle L_1: \frac{x - 9}{4} = \frac{y + 2}{-3} = \frac{z}{1}$ and $\displaystyle L_2: \frac{x}{-2} = \frac{y + 7}{9} = \frac{z - 2}{2}$.

  • Find the equation of the plane that passes through the origin and the point $(6, -3, 2)$, and is perpendicular to the plane $4x - y + 2z = 8$.

  • There are two intersection planes $x - y + z = 7$ and $3x + 2y - 12z + 5 = 0$,
    1. Find the intersecting angle between the two planes
    2. Find a plane that is through the point $P(1,1,1)$ and orthogonal to both of the two planes.

  • Find a plane that is through the points $P(1,1, - 1)$, $Q(3,0,2)$, and $R(-2,1,0)$.

  • Let the lines $\dfrac{x + 2}{1} = \dfrac{y - 3}{-1} = \dfrac{z + 1}{1}$, and $\dfrac{x + 4}{2} = \dfrac{y}{1} = \dfrac{z - 4}{3}$. Find the equation of the line perpendicular to and intersects both lines.

  • Find the distance between the line $L_1: \begin{cases} x - y = 0, \ z = 0 \end{cases}$ and the line $\displaystyle L_2: \frac{x - 2}{4} = \frac{y - 1}{-2} = \frac{z - 3}{-1}$.

  • Find the equation of a plane that passes through the point $(-1, 3, 1)$ and is perpendicular to each of the two mutually perpendicular planes $\begin{cases} 2x + y - 3z + 2 = 0 \ 5x + 4y - 3z + 4 = 0 \end{cases}$.

  • Find the equation of the plane that passes through the line $\dfrac{x - 1}{2} = \dfrac{y + 2}{-3} = \dfrac{z - 2}{2}$ and is perpendicular to the plane $3x + 2y - z - 5 = 0$.

  • Do the line $\dfrac{x - 1}{1} = \dfrac{y + 1}{2} = \dfrac{z - 1}{\lambda}$ and the line $x + 1 = y - 1 = z$ intersect? If they intersect, find their intersection point. If not, find the distance between the two lines.

  • Find the distance between the parallel planes $10x + 2y - 2z = 5$ and $5x + y - z = 1$.

  • Find the equation of the plane that passes through the intersection of the hyperboloid of one sheet $\displaystyle \frac{x^{2}}{4} + \frac{y^{2}}{2} - 2z^{2} = 1$ and the sphere $x^{2} + y^{2} + z^{2} = 4$, and is perpendicular to the line $\begin{cases} x = 0, \ 3y + z = 0 \end{cases}$.