Vector Algebra
Resultant of Vectors
Grade 12
Question:
<p><i>ABC</i> is a triangle, right angled at <i>A</i>. The resultant of the forces acting along \(\overrightarrow{AB}\), \(\overrightarrow{AC}\) with magnitudes \(1/AB\) and \(1/AC\) respectively is the force along \(\overrightarrow{AD}\), where <i>D</i> is the foot of the perpendicular from <i>A</i> onto <i>BC</i>. The magnitude of the resultant is</p>
<p>\(\dfrac{AB^2 + AC^2}{(AB)^2(AC)^2}\)</p>
<p>\(\dfrac{(AB)+(AC)}{AB+AC}\)</p>
<p>\(\dfrac{1}{AB} + \dfrac{1}{AC}\)</p>
<p>\(\dfrac{1}{AD}\)</p>
Step-by-Step Solution
Key Concept: The resultant of two perpendicular forces with magnitudes 1/AB and 1/AC can be found using vector addition. The key is recognizing that AD is perpendicular to BC, and using the geometric relationship in a right triangle where 1/AD² = 1/AB² + 1/AC² (altitude-on-hypotenuse formula).
Step 1: Set up coordinates with A at origin, B on positive x-axis at distance c = AB, and C on positive y-axis at distance b = AC. Step 2: The forces are: F_1 = (1/c)î (along AB) and F_2 = (1/b)ĵ (along AC). Resultant: R = (1/c)î + (1/b)ĵ Step 3: For a right triangle with right angle at A, the altitude from A to hypotenuse BC has length h = AD = (bc)/√(b^2 + c^2), where the hypotenuse BC = √(b^2 + c^2). Step 4: The direction along AD makes an angle θ with x-axis where tan(θ) = b/c (same as slope of altitude). For R to point along AD: (1/b)/(1/c) = b/c, which is satisfied. Step 5: Magnitude of R: |R| = √((1/c)^2 + (1/b)^2) = √((b^2 + c^2)/(b^2c^2)) = √(b^2 + c^2)/(bc) = 1/h = 1/AD ∴ Answer: D
Correct Answer: D