Vector Algebra
Applications of Vector Products
Grade 12
Question:
<p>The angles of a triangle, two of whose sides are represented by vectors <strong>3(a × b)</strong> and <strong>b - (a · b)a</strong>, where <strong>b</strong> is a non-zero vector and <strong>a</strong> is a unit vector in the direction of <strong>a</strong>, are</p>
<p>(a) <strong>tan</strong><sup>-1</sup><strong>(</strong>√3<strong>)</strong></p>
<p>(b) <strong>tan</strong><sup>-1</sup><strong>(1/</strong>√3<strong>)</strong></p>
<p>(c) <strong>cot</strong><sup>-1</sup><strong>(0)</strong></p>
<p>(d) <strong>tan</strong><sup>-1</sup><strong>(1)</strong></p>
Step-by-Step Solution
Key Concept: Recognize that the two vectors are perpendicular and use the sine rule to find the remaining angles of the triangle.
Step 1: Let v_1 = 3(a × b) and v_2 = b - (a · b)a . These represent two sides of the triangle. Step 2: Check if v_1 · v_2 = 0 : The two sides are perpendicular, so one angle is 90°. Step 3: Using the sine rule: tan θ = (1/3)|b - (a · b)a| / |a × b| Step 4: Simplify using |(a × b) × a| = |a × b| |a| sin 90° = |a × b| Step 5: This gives tan θ = 1/√3 , so θ = π/6 . Step 6: The third angle is π - 90° - 30° = 60° , which gives tan<sup>-1</sup>(√3) . ∴ Answer is (a).
Correct Answer: a