Q3 If θ is the angle, in degrees, between the longest diagonal of the cube and one of the edges of the cube, then cos θ =
| (A) | 1/2 |
|---|---|
| (B) | 1/√2 |
| (C) | √2 / 3 |
| (D) | 1 / √3 |
Problem Understanding
The angle θ between longest face diagonal and space diagonal of the cube. One edge has length a = 1 for simplicity. Space diagonal vector from (0,0,0) to (1,1,1), magnitude d = √3
Cosine formula gives:
cos θ = (�vec{d} • �vec{r}) / (|d| |r|) = (1·1 + 1·0 + 1·0) / (√3 · 1) = 1/√3 ≈ 0.577
Geometric: Right triangle with space diagonal (hypotenuse |d| = √3), face diagonal (|f| = √2), edge. Thus cos θ = adjacent/hypotenuse = 1/√3
The cosine of angle between cube space diagonal and edge is a fundamental geometry problem frequently appearing in CSIR NET Life Sciences, GATE, and competitive exams. This cube diagonal edge angle calculation uses vector dot products or Pythagorean theorem in 3D space.
Why This Matters for CSIR NET
Cube geometry questions test spatial reasoning and vector applications, crucial for molecular modeling and crystal structure analysis in life sciences. The longest diagonal of cube (body diagonal a✓3) forms angle θ with edges where cosθ=1/✓3.
Step-by-Step Derivation
- Space diagonal vector: d = (a,a,a), length a✓3.
- Edge vector: e = (a,0,0), length a.
- Dot product: d⋅e = a2
- cos θ = a2 / (a✓3 ⋅ a) = 1/✓3.
Keywords: cube longest diagonal edge angle, cos theta cube diagonal, space diagonal cube cosine, CSIR NET cube geometry.
