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The vector a = (1, 3, sin 2α) forms an angle greater than 90° with the z-axis. Given that 2α is negative, and vectors b and c are perpendicular, b ⋅ c equals zero. The equation tan² α − tan α − 6 = 0 has solutions tan α = 3 or −2. If tan α = 3, then sin 2α = 2 tan α / (1 + tan² α) = 3/5, which is positive and contradicts the condition (2α < 0). Thus, tan α = −2. For this value, sin 2α = 2 tan α / (1 − tan² α) = 4/3, which is greater than zero. However, sin 2α must be negative, placing 2α in the third quadrant and α/2 in the first quadrant. The square root of sin(α/2) is valid, and α is given by (4n + 1)π − tan⁻¹ 2.

1. The vector a = (1, 3, sin 2α) forms an angle greater than 90° with the z-axis.
2. The equation tan² α − tan α − 6 = 0 has solutions tan α = 3 or −2.
3. The value of sin 2α is calculated as 2 tan α / (1 + tan² α) = 3/5, which is positive.
4. The angle α is expressed as (4n + 1)π − tan⁻¹ 2.

Correct answer: The vector a = (1, 3, sin 2α) forms an angle greater than 90° with the z-axis.

Solution

For the vector a = (1, 3, sin 2α) to form an angle greater than 90° with the z-axis, sin 2α must be negative. Given 2α < 0, the correct quadrant for sin 2α ensures this condition is satisfied.

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