<p>If <strong>u</strong> and <strong>v</strong> are unit vectors and θ is the acute angle between them, then 2<strong>u</strong> × 3<strong>v</strong> is a unit vector for</p>
<p>(a) exactly two values of θ</p>
<p>(b) more than two values of θ</p>
<p>(c) no value of θ</p>
<p>(d) exactly one value of θ</p>
Step-by-Step Solution
Key Concept: The magnitude of a cross product of two vectors equals the product of their magnitudes times the sine of the angle between them. For 2u × 3v to be a unit vector, we need |2u × 3v| = 1, which gives us a specific trigonometric equation in θ.
Step 1: Write the magnitude formula for the cross product.
Given: u and v are unit vectors, so |u| = 1 and |v| = 1.
We need |2u × 3v| = 1 (for it to be a unit vector). Step 2: Apply the cross product magnitude property.
|2u × 3v| = |2| · |u| × |3| · |v| · sin(θ)
= 2 · 1 · 3 · 1 · sin(θ)
= 6sin(θ) Step 3: Set up the equation for unit vector condition.
For 2u × 3v to be a unit vector:
6sin(θ) = 1
sin(θ) = 1/6 Step 4: Determine the number of solutions in the acute angle range.
Since θ is acute, we need θ ∈ (0, π/2).
In this range, sin(θ) is strictly increasing from 0 to 1.
Since 0 < 1/6 < 1, the equation sin(θ) = 1/6 has exactly one solution in (0, π/2).
This solution is θ = arcsin(1/6). ∴ Answer: D
Correct Answer: D