Find the angle between a body diagonal of a cube and any one of its face diagonals. [Hint: Choose a cube with side 1 and with one corner at 0 and the opposite corner at the point (1, 1, 1). Write down the vector that represents a body diagonal and another that represents a face diagonal, and then find the angle between them as in Problem 1.4.]
Problem 1.4.
One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors b = (1, 2, 4) and c = (4, 2, 1) by evaluating their scalar product.
The correct answer and explanation is:
Let’s solve this step by step.
Step 1: Define the Vectors
Consider a cube with side length 1. Let one vertex be at the origin (0,0,0)(0,0,0) and the opposite vertex at (1,1,1)(1,1,1).
Body Diagonal:
A body diagonal of the cube runs from (0,0,0)(0,0,0) to (1,1,1)(1,1,1), so the vector representing it is: d=(1,1,1)\mathbf{d} = (1,1,1)
Face Diagonal:
A face diagonal lies on one of the cube’s faces. For example, the diagonal of the face in the xyxy-plane from (0,0,0)(0,0,0) to (1,1,0)(1,1,0) is: f=(1,1,0)\mathbf{f} = (1,1,0)
Step 2: Compute the Dot Product
The dot product of two vectors d\mathbf{d} and f\mathbf{f} is given by: d⋅f=(1⋅1)+(1⋅1)+(1⋅0)=1+1+0=2\mathbf{d} \cdot \mathbf{f} = (1 \cdot 1) + (1 \cdot 1) + (1 \cdot 0) = 1 + 1 + 0 = 2
Step 3: Compute the Magnitudes
The magnitudes of the vectors are: ∣d∣=12+12+12=3|\mathbf{d}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} ∣f∣=12+12+02=2|\mathbf{f}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}
Step 4: Compute the Angle
Using the formula for the angle between two vectors: cosθ=d⋅f∣d∣∣f∣\cos \theta = \frac{\mathbf{d} \cdot \mathbf{f}}{|\mathbf{d}| |\mathbf{f}|} cosθ=23⋅2=26=63\cos \theta = \frac{2}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}
Taking the inverse cosine: θ=cos−1(63)\theta = \cos^{-1} \left(\frac{\sqrt{6}}{3}\right)
Explanation
The problem involves understanding vector relationships in a cube. The body diagonal connects two opposite corners, while a face diagonal connects two opposite points on a face. By choosing a coordinate system where one corner is at the origin, we can easily express these vectors.
Using the dot product, we find their similarity. Since both vectors lie in three-dimensional space, we compute their magnitudes and apply the formula for the cosine of the angle between them. This technique is fundamental in physics, engineering, and computer graphics, where vector angles determine forces, movements, and perspectives.
