Understanding the Cuboid and Vertices

The problem provides the coordinates of three vertices of the base (A, B, C) and one vertex of the top face (G) of a cuboid ABCDEFGH. The base is ABCD. This means the edges AB, BC, and AD are along the axes starting from A if A is at the origin.

Given vertices:

• A = (0, 0, 0)
• B = (12, 0, 0)
• C = (12, 6, 0)
• G = (12, 6, 4)

From these coordinates, we can deduce the dimensions of the cuboid and the coordinates of other points if needed. Since A is the origin (0, 0, 0) and B is (12, 0, 0), the length of AB is 12 units along the x-axis. Since C is (12, 6, 0), it is in the xy-plane, 12 units along x and 6 units along y from A (relative to the plane). Thus, BC is parallel to the y-axis with length 6. The z-coordinate represents the height. Since G is (12, 6, 4), which corresponds to vertex C (12, 6, 0) but with a height of 4, the height of the cuboid is 4 units. This confirms the structure is a cuboid with dimensions 12 x 6 x 4.

Calculating Vectors for Angle Measurement

To find the angles between line segments originating from A, we need to find the vectors corresponding to these segments. The vectors are obtained by subtracting the coordinates of the initial point (A) from the coordinates of the terminal point.

• Vector $\vec{AB}$: This vector goes from A to B. $\vec{AB} = B - A = (12 - 0, 0 - 0, 0 - 0) = (12, 0, 0)$.
• Vector $\vec{AC}$: This vector goes from A to C. $\vec{AC} = C - A = (12 - 0, 6 - 0, 0 - 0) = (12, 6, 0)$.
• Vector $\vec{AG}$: This vector goes from A to G. $\vec{AG} = G - A = (12 - 0, 6 - 0, 4 - 0) = (12, 6, 4)$.

Finding cos α using the Dot Product

The angle $\alpha$ is the angle between $\vec{AB}$ and $\vec{AG}$. We can find the cosine of this angle using the dot product formula:

$\cos \alpha = \frac{\vec{AB} \cdot \vec{AG}}{|\vec{AB}| |\vec{AG}|}$

First, calculate the dot product $\vec{AB} \cdot \vec{AG}$:

$\vec{AB} \cdot \vec{AG} = (12)(12) + (0)(6) + (0)(4) = 144 + 0 + 0 = 144$

Next, calculate the magnitudes of the vectors $|\vec{AB}|$ and $|\vec{AG}|$.

$|\vec{AB}| = \sqrt{12^2 + 0^2 + 0^2} = \sqrt{144} = 12$

$|\vec{AG}| = \sqrt{12^2 + 6^2 + 4^2} = \sqrt{144 + 36 + 16} = \sqrt{196} = 14$

Now, substitute these values into the cosine formula for $\alpha$:

$\cos \alpha = \frac{144}{12 \times 14} = \frac{144}{168}$

Simplify the fraction:

$\cos \alpha = \frac{144 \div 24}{168 \div 24} = \frac{6}{7}$

Finding cos β using the Dot Product

The angle $\beta$ is the angle between $\vec{AC}$ and $\vec{AG}$. We use the same dot product formula:

$\cos \beta = \frac{\vec{AC} \cdot \vec{AG}}{|\vec{AC}| |\vec{AG}|}$

First, calculate the dot product $\vec{AC} \cdot \vec{AG}$:

$\vec{AC} \cdot \vec{AG} = (12)(12) + (6)(6) + (0)(4) = 144 + 36 + 0 = 180$

Next, calculate the magnitudes of the vectors $|\vec{AC}|$ and $|\vec{AG}|$. We already found $|\vec{AG}| = 14$.

$|\vec{AC}| = \sqrt{12^2 + 6^2 + 0^2} = \sqrt{144 + 36} = \sqrt{180}$

Simplify the magnitude $|\vec{AC}|$: $\sqrt{180} = \sqrt{36 \times 5} = \sqrt{36} \times \sqrt{5} = 6\sqrt{5}$.

Now, substitute these values into the cosine formula for $\beta$:

$\cos \beta = \frac{180}{6\sqrt{5} \times 14} = \frac{180}{84\sqrt{5}}$

Simplify the fraction and rationalize the denominator:

$\cos \beta = \frac{180}{84\sqrt{5}} = \frac{180 \div 12}{84 \div 12}\frac{1}{\sqrt{5}} = \frac{15}{7\sqrt{5}} = \frac{15\sqrt{5}}{7\sqrt{5} \times \sqrt{5}} = \frac{15\sqrt{5}}{7 \times 5} = \frac{15\sqrt{5}}{35} = \frac{3\sqrt{5}}{7}$

Calculating cos 2α and cos 2β

The question asks for the value of $\cos 2\alpha + \cos 2\beta$. We can use the double angle trigonometric identity: $\cos 2\theta = 2\cos^2\theta - 1$.

For $\cos 2\alpha$, using $\cos \alpha = \frac{6}{7}$:

$\cos 2\alpha = 2\cos^2\alpha - 1 = 2\left(\frac{6}{7}\right)^2 - 1 = 2\left(\frac{36}{49}\right) - 1 = \frac{72}{49} - 1 = \frac{72 - 49}{49} = \frac{23}{49}$

For $\cos 2\beta$, using $\cos \beta = \frac{3\sqrt{5}}{7}$:

$\cos 2\beta = 2\cos^2\beta - 1 = 2\left(\frac{3\sqrt{5}}{7}\right)^2 - 1 = 2\left(\frac{9 \times 5}{49}\right) - 1 = 2\left(\frac{45}{49}\right) - 1 = \frac{90}{49} - 1 = \frac{90 - 49}{49} = \frac{41}{49}$

Final Calculation: cos 2α + cos 2β

Now, we add the values of $\cos 2\alpha$ and $\cos 2\beta$:

$\cos 2\alpha + \cos 2\beta = \frac{23}{49} + \frac{41}{49}$

Since the fractions have a common denominator, we simply add the numerators:

$\cos 2\alpha + \cos 2\beta = \frac{23 + 41}{49} = \frac{64}{49}$

Revision Table: Key Calculations

Additional Information: Vectors and Angles in 3D Geometry

This problem involves several key concepts from 3D coordinate geometry and vector algebra. Understanding these concepts is crucial for solving problems related to points, lines, and planes in space.

• Vectors from Coordinates: A vector from point P($x_1, y_1, z_1$) to point Q($x_2, y_2, z_2$) is given by $\vec{PQ} = (x_2 - x_1, y_2 - y_1, z_2 - z_1)$. This represents the displacement from P to Q.
• Magnitude of a Vector: The magnitude (or length) of a vector $\vec{v} = (v_x, v_y, v_z)$ is given by $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$. This is the distance from the origin to the point $(v_x, v_y, v_z)$ if the vector starts at the origin, or simply the length of the line segment representing the vector.
• Dot Product: The dot product of two vectors $\vec{u} = (u_x, u_y, u_z)$ and $\vec{v} = (v_x, v_y, v_z)$ is a scalar quantity defined as $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$. The dot product is related to the angle between the vectors.
• Angle between Two Vectors: The cosine of the angle $\theta$ between two non-zero vectors $\vec{u}$ and $\vec{v}$ is given by the formula $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$. This formula is derived directly from the definition of the dot product.
• Trigonometric Identities: The double angle formula $\cos 2\theta = 2\cos^2\theta - 1$ (or $1 - 2\sin^2\theta$ or $\cos^2\theta - \sin^2\theta$) is essential for relating the cosine of double an angle to the cosine (or sine) of the angle itself.
• Cuboid Geometry: A cuboid is a 3D solid with six rectangular faces. Its vertices can be represented using coordinates based on its dimensions. If one vertex is at the origin and the edges are aligned with the axes, the vertices can be easily determined.