13.4 Using Spatial Vectors to Compute Angles and Distances

13.4 Using Spatial Vectors to Compute Angles and Distances

In the Gaokao Mathematics examination, Section II of solid geometry questions carries 8 to 12 marks and routinely tests vector calculations of spatial angles and distances. Using spatial vector direction vectors and plane normal vectors transforms visual geometric intuition into systematic algebraic computation.

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1. Vector Formulas for Spatial Angles

1. Angle Between Two Lines (异面直线或相交直线所成角)

Let $l_1$ and $l_2$ be two straight lines in space with direction vectors $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$, respectively. Let $\theta$ be the angle between $l_1$ and $l_2$ (where $\theta \in (0, \pi/2]$): $\cos\theta = |\cos \langle \boldsymbol{v}_1, \boldsymbol{v}_2 \rangle| = \frac{|\boldsymbol{v}_1 \cdot \boldsymbol{v}_2|}{|\boldsymbol{v}_1||\boldsymbol{v}_2|}$ Note: The absolute value in the numerator ensures $\cos\theta \ge 0$, reflecting the acute/right angle convention for line angles.

2. Angle Between a Line and a Plane (线面角)

Let line $l$ have direction vector $\boldsymbol{v}$, and plane $\alpha$ have normal vector $\boldsymbol{n}$. Let $\theta$ be the angle between line $l$ and plane $\alpha$ (where $\theta \in [0, \pi/2]$): $\sin\theta = |\cos \langle \boldsymbol{v}, \boldsymbol{n} \rangle| = \frac{|\boldsymbol{v} \cdot \boldsymbol{n}|}{|\boldsymbol{v}||\boldsymbol{n}|}$ Mathematical Intuition: The angle between line direction vector $\boldsymbol{v}$ and plane normal vector $\boldsymbol{n}$ is complementary to the line-plane angle $\theta$, so $\cos(90^\circ - \theta) = \sin\theta$.

3. Dihedral Angle Between Two Planes (二面角)

Let planes $\alpha$ and $\beta$ have normal vectors $\boldsymbol{n}_1$ and $\boldsymbol{n}_2$, respectively. Let $\theta$ be the dihedral angle $\alpha-l-\beta$ (where $\theta \in [0, \pi]$): $|\cos\theta| = |\cos \langle \boldsymbol{n}_1, \boldsymbol{n}_2 \rangle| = \frac{|\boldsymbol{n}_1 \cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|}$ Determining Acute vs. Obtuse Dihedral Angle:

• Inspect the geometric diagram: If the dihedral angle is visibly acute, then $\cos\theta = \frac{|\boldsymbol{n}_1 \cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|} > 0$.
• If the dihedral angle is visibly obtuse, then $\cos\theta = -\frac{|\boldsymbol{n}_1 \cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_1||\boldsymbol{n}_2|} < 0$.
• Direction rule: If both normal vectors $\boldsymbol{n}_1, \boldsymbol{n}_2$ point into the interior of the dihedral angle (or both point outward), then the dihedral angle is supplementary to $\langle \boldsymbol{n}_1, \boldsymbol{n}_2 \rangle$.

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2. Vector Formulas for Spatial Distances

1. Distance From a Point to a Plane (点到平面的距离)

Let $P$ be a point in space, and plane $\alpha$ pass through point $A$ with normal vector $\boldsymbol{n}$. The distance $d$ from point $P$ to plane $\alpha$ is the magnitude of the projection of vector $\vec{AP}$ along $\boldsymbol{n}$: $d = \frac{|\vec{AP} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}$

2. Distance Between Parallel Planes (平行平面间的距离)

To find the distance between parallel planes $\alpha \parallel \beta$, choose any arbitrary point $P \in \alpha$, and compute the distance from $P$ to plane $\beta$ using the point-to-plane distance formula.

3. Distance From a Point to a Line (点到直线的距离)

Let $P$ be a point, and line $l$ pass through point $A$ with direction vector $\boldsymbol{v}$. The distance $d$ from $P$ to line $l$ is: $d = \sqrt{|\vec{AP}|^2 - \left( \frac{\vec{AP} \cdot \boldsymbol{v}}{|\boldsymbol{v}|} \right)^2}$

4. Distance Between Two Skew Lines (异面直线间的距离)

Let skew lines $l_1, l_2$ pass through points $A_1, A_2$ with direction vectors $\boldsymbol{v}_1, \boldsymbol{v}_2$. Let $\boldsymbol{n}$ be a common normal vector perpendicular to both lines ($\boldsymbol{n} = \boldsymbol{v}_1 \times \boldsymbol{v}_2$). The distance $d$ between $l_1$ and $l_2$ is: $d = \frac{|\vec{A_1 A_2} \cdot \boldsymbol{n}|}{|\boldsymbol{n}|}$

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3. Angle and Distance Formula Summary Table

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4. Classic 12-Mark Gaokao Worked Example

Problem

In a right trihedral prism $ABC-A_1B_1C_1$, the base $ABC$ is an equilateral triangle with side length $2$. The height of the prism is $AA_1 = 3$. Point $D$ is the midpoint of edge $A_1C_1$.

1. Prove that line $B_1D \perp$ plane $ACC_1A_1$.
2. Calculate the cosine of the dihedral angle $A-B_1D-C$.
3. Calculate the distance from point $A_1$ to plane $B_1DC$.

Solution

Step 1: Coordinate System Setup

Let $O$ be the midpoint of base edge $AC$. In equilateral $\triangle ABC$, $BO \perp AC$ and $BO = \sqrt{3}$.

• Take $O$ as the origin $(0,0,0)$.
• Let $x$-axis be along directed line $OC$.
• Let $y$-axis be along directed line $OB$.
• Let $z$-axis be parallel to vertical edge $AA_1$.

Coordinates of key points:

• Base vertices: $O(0,0,0)$, $A(-1, 0, 0)$, $C(1, 0, 0)$, $B(0, \sqrt{3}, 0)$.
• Top vertices ($z=3$): $A_1(-1, 0, 3)$, $C_1(1, 0, 3)$, $B_1(0, \sqrt{3}, 3)$.
• Midpoint $D$ of $A_1C_1$: $D(0, 0, 3)$.

Step 2: Part 1 - Vector Proof of Line-Plane Perpendicularity

• Direction vector of $B_1D$: $\vec{B_1D} = D - B_1 = (0-0, 0-\sqrt{3}, 3-3) = (0, -\sqrt{3}, 0)$
• Plane $ACC_1A_1$ is the $xz$-plane ($y=0$). Two non-collinear vectors in $ACC_1A_1$ are: $\vec{AC} = C - A = (2, 0, 0), \quad \vec{AA_1} = A_1 - A = (0, 0, 3)$
• Test dot products: $\vec{B_1D} \cdot \vec{AC} = 0(2) + (-\sqrt{3})(0) + 0(0) = 0$ $\vec{B_1D} \cdot \vec{AA_1} = 0(0) + (-\sqrt{3})(0) + 0(3) = 0$
• Since $\vec{B_1D}$ is perpendicular to two intersecting vectors in plane $ACC_1A_1$, line $B_1D \perp$ plane $ACC_1A_1$. (Part 1 Q.E.D.)

Step 3: Part 2 - Calculate Dihedral Angle $A-B_1D-C$

We find normal vectors for plane $AB_1D$ and plane $CB_1D$.

• Normal vector $\boldsymbol{n}_1$ of plane $AB_1D$: Vectors in $AB_1D$: $\vec{AD} = D - A = (1, 0, 3)$ and $\vec{B_1D} = (0, -\sqrt{3}, 0)$. $\begin{cases} \boldsymbol{n}_1 \cdot \vec{B_1D} = 0 \implies -\sqrt{3}y = 0 \implies y = 0 \\ \boldsymbol{n}_1 \cdot \vec{AD} = 0 \implies x + 3z = 0 \implies x = -3z \end{cases}$ Let $z=1 \implies x=-3, y=0$. So $\boldsymbol{n}_1 = (-3, 0, 1)$. Magnitude $|\boldsymbol{n}_1| = \sqrt{9+0+1} = \sqrt{10}$.

• Normal vector $\boldsymbol{n}_2$ of plane $CB_1D$: Vectors in $CB_1D$: $\vec{CD} = D - C = (-1, 0, 3)$ and $\vec{B_1D} = (0, -\sqrt{3}, 0)$. $\begin{cases} \boldsymbol{n}_2 \cdot \vec{B_1D} = 0 \implies y = 0 \\ \boldsymbol{n}_2 \cdot \vec{CD} = 0 \implies -x + 3z = 0 \implies x = 3z \end{cases}$ Let $z=1 \implies x=3, y=0$. So $\boldsymbol{n}_2 = (3, 0, 1)$. Magnitude $|\boldsymbol{n}_2| = \sqrt{9+0+1} = \sqrt{10}$.

• Cosine of Dihedral Angle: $\cos\langle \boldsymbol{n}_1, \boldsymbol{n}_2 \rangle = \frac{\boldsymbol{n}_1 \cdot \boldsymbol{n}_2}{|\boldsymbol{n}_1||\boldsymbol{n}_2|} = \frac{(-3)(3) + 0(0) + (1)(1)}{\sqrt{10}\sqrt{10}} = \frac{-9 + 1}{10} = -\frac{8}{10} = -\frac{4}{5}$ Observing the figure, the dihedral angle $A-B_1D-C$ is obtuse, so its cosine is negative: $\cos\theta = -\frac{4}{5}$

Step 4: Part 3 - Point-to-Plane Distance Calculation

• Compute distance from $A_1(-1, 0, 3)$ to plane $B_1DC$ (normal vector $\boldsymbol{n}_2 = (3, 0, 1)$):
• Choose point $D(0, 0, 3) \in$ plane $B_1DC$.
• Vector $\vec{DA_1} = A_1 - D = (-1-0, 0-0, 3-3) = (-1, 0, 0)$.
• Apply point-to-plane distance formula: $d = \frac{|\vec{DA_1} \cdot \boldsymbol{n}_2|}{|\boldsymbol{n}_2|} = \frac{|(-1)(3) + 0(0) + 0(1)|}{\sqrt{3^2 + 0^2 + 1^2}} = \frac{|-3|}{\sqrt{10}} = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$
• Conclusion: The distance from point $A_1$ to plane $B_1DC$ is $\frac{3\sqrt{10}}{10}$.