Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>Let \ \(\vec{a}\), \ \(\vec{b}\), \ \(\vec{c}\) \ be three vectors and \ \(Y\) \ be the surface area defined as:</p><p>\[Y = 2\left(\frac{|\vec{b} \times \vec{c}|}{[\vec{a}\,\vec{b}\,\vec{c}]} + \frac{|\vec{c} \times \vec{a}|}{[\vec{a}\,\vec{b}\,\vec{c}]} + \frac{|\vec{a} \times \vec{b}|}{[\vec{a}\,\vec{b}\,\vec{c}]}\right)\]</p><p>If \ \(Y = 4\), find \ \(\cos\alpha\) \ and the angle \ \(\alpha\).</p>
<p>\(\cos\alpha = \dfrac{1}{2}\)</p>
<p>\(\alpha = 60^\circ\)</p>
<p>\(\alpha = 45^\circ\)</p>
<p>\(\cos\alpha = \dfrac{1}{\sqrt{2}}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the expression represents twice the sum of reciprocals of altitudes of a tetrahedron formed by vectors a, b, c. The scalar triple product [a b c] equals the volume, and each cross product magnitude divided by volume gives the reciprocal of the corresponding altitude.
Step 1: Interpret the surface area formula. For a tetrahedron with vectors a, b, c: • |b × c|/[a b c] = 1/h_1 (reciprocal of altitude to face bc) • |c × a|/[a b c] = 1/h_2 (reciprocal of altitude to face ca) • |a × b|/[a b c] = 1/h_3 (reciprocal of altitude to face ab) Step 2: Given Y = 4, we have: 2(1/h_1 + 1/h_2 + 1/h_3) = 4 ∴ 1/h_1 + 1/h_2 + 1/h_3 = 2 Step 3: For a regular tetrahedron, h_1 = h_2 = h_3 = h: 3/h = 2 → h = 3/2 Step 4: In a regular tetrahedron, the angle α between any two edges from the same vertex satisfies: cos α = 1/3 (derived from edge geometry where opposite edges are perpendicular) ∴ α = arccos(1/3) ≈ 70.53° or cos α = 1/3
Correct Answer: A