Vector Algebra
Dot product and orthogonality
Grade 12
Question:
<p>Vectors <strong>b</strong> = (tan α, −1, 2 sin α/2) and <strong>c</strong> = (tan α, tan α, −3 sin α/2) are orthogonal and vectors <strong>a</strong> = (1, 3, sin 2α) makes an obtuse angle with the Z-axis, then the value of α is</p>
<p>(a) α = (4n + 1)π + tan⁻¹ 2</p>
<p>(b) α = (4n + 1)π − tan⁻¹ 2</p>
<p>(c) α = (4n + 2)π + tan⁻¹ 2</p>
<p>(d) α = (4n + 2)π − tan⁻¹ 2</p>
Step-by-Step Solution
Key Concept: Use orthogonality condition to relate tan α to the angle constraints, then verify which solution satisfies the obtuse angle requirement with the z-axis.
Step 1: Since a = (1, 3, sin 2α) makes an obtuse angle with the Z-axis, its z-component is negative: \(-1 ≤ \sin 2α < 0\) ...(i) Step 2: Given b ⊥ c , so b · c = 0: \(\tan α · \tan α + (-1) · \tan α + 2\sin(α/2) · (-3\sin(α/2)) = 0\) \(\tan^2 α - \tan α - 6\sin^2(α/2) = 0\) Step 3: Solving: \((\tan α - 3)(\tan α + 2) = 0\) \(\tan α = 3 \text{ or } \tan α = -2\) Step 4: If \(\tan α = 3\): \(\sin 2α = \frac{2\tan α}{1 + \tan^2 α} = \frac{6}{10} = \frac{3}{5} > 0\) (not possible) Step 5: If \(\tan α = -2\): \(\sin 2α = \frac{2(-2)}{1 + 4} = \frac{-4}{5} < 0\) (satisfies condition i) Step 6: Therefore \(\tan α = -2\), which gives: \(α = (4n + 1)π - \tan^{-1} 2 \text{ or } α = (4n + 2)π - \tan^{-1} 2\) ∴ Answer is (b) and (d).
Correct Answer: B, D