Vector Algebra
Vector Magnitude
Grade 12
Question:
<p>Let \(\vec{a}\) and \(\vec{b}\) be two unit vectors and \(\theta\) is the angle between them. Then \(\vec{a} + \vec{b}\) is a unit vector, if</p>
<p>(a) \(\theta = \frac{\pi}{4}\)</p>
<p>(b) \(\theta = \frac{\pi}{3}\)</p>
<p>(c) \(\theta = \frac{\pi}{2}\)</p>
<p>(d) \(\theta = \frac{2\pi}{3}\)</p>
Step-by-Step Solution
Key Concept: Use the magnitude formula for vector addition: |a + b|² = |a|² + |b|² + 2|a||b|cos(θ). Since a and b are unit vectors and a + b is also a unit vector, we can set up an equation to solve for θ.
Step 1: Given that |a| = 1 and |b| = 1 (unit vectors), and |a + b| = 1 (a + b is also a unit vector). Step 2: Use the formula for magnitude of vector sum: |a + b|^2 = |a|^2 + |b|^2 + 2|a||b|cos(θ) Step 3: Substitute the known values: 1^2 = 1^2 + 1^2 + 2(1)(1)cos(θ) 1 = 1 + 1 + 2cos(θ) Step 4: Simplify the equation: 1 = 2 + 2cos(θ) -1 = 2cos(θ) cos(θ) = -1/2 Step 5: Find θ in the range [0, π]: θ = arccos(-1/2) = 2π/3 Step 6: Verify: When θ = 2π/3, we have cos(2π/3) = -1/2, which gives |a + b|^2 = 1 + 1 + 2(-1/2) = 2 - 1 = 1, so |a + b| = 1 ✓ ∴ Answer: D
Correct Answer: D