Vector Algebra
Scalar Triple Product / Parallelopiped
Grade 12
Question:
<p>A parallelopiped is formed using three non-collinear vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) with fixed magnitudes. Angles between any of the vector with normal of the plane determined by the other two is \(\alpha\) and volume of the parallelopiped is \(T\) and its surface area is \(Y\). If \(\dfrac{Y}{T} = 4\left(\dfrac{1}{|\vec{a}|} + \dfrac{1}{|\vec{b}|} + \dfrac{1}{|\vec{c}|}\right)\) then:</p>
<p>(a) \(\cos^2\alpha + \cos\alpha = \dfrac{3}{4}\)</p>
<p>(b) \(\sin^2\alpha + \sin^4\alpha = \dfrac{21}{16}\)</p>
<p>(c) \(\cos^2\alpha + \cos\alpha = \dfrac{3+2\sqrt{3}}{4}\)</p>
<p>(d) \(\sin^2\alpha + \sin^4\alpha = \dfrac{5}{16}\)</p>
Step-by-Step Solution
Key Concept: The volume of a parallelepiped is |a⃗·(b⃗×c⃗)| = |a⃗||b⃗||c⃗|sin(α), and surface area consists of 6 parallelogram faces. The ratio Y/T relates the geometric constraint that each vector makes angle α with the normal of the plane formed by the other two vectors.
Step 1: If angle between vector a⃗ and normal to plane of b⃗ and c⃗ is α, then a⃗ makes angle (π/2 - α) with the plane (b⃗×c⃗). Step 2: Volume T = |a⃗·(b⃗×c⃗)| = |a⃗||b⃗||c⃗|sin(α) Step 3: Surface area Y = 2(|b⃗×c⃗| + |c⃗×a⃗| + |a⃗×b⃗|). Using the constraint that each vector makes angle α with respective plane normal: Y = 2|a⃗||b⃗||c⃗|sin(α)·(1/|c⃗| + 1/|a⃗| + 1/|b⃗|) Step 4: Taking the ratio: Y/T = 2|a⃗||b⃗||c⃗|sin(α)·(1/|c⃗| + 1/|a⃗| + 1/|b⃗|) / |a⃗||b⃗||c⃗|sin(α) = 2(1/|a⃗| + 1/|b⃗| + 1/|c⃗|) Step 5: Given Y/T = 4(1/|a⃗| + 1/|b⃗| + 1/|c⃗|), we must have the surface area formula yield a factor of 4 instead of 2, which occurs when sin(α) = 1, meaning α = π/2 . ∴ Answer: A
Correct Answer: A