If θ is the angle between two vectors a⃗ and b⃗, then a⃗ ⋅ b⃗ ≥ 0 only when

B — Option B is correct because the dot product formula states that a⃗ ⋅ b⃗ = a⃗ b⃗ cos θ. For the dot product to be non-negative (≥ 0), we need cos θ ≥ 0 since magnitudes a⃗ and b⃗ are always non-negative. The cosine function is non-negative when 0 ≤ θ ≤ π/2. At θ = 0, cos θ = 1, giving maximum positive value. At θ = π/2, cos θ = 0, making the dot product exactly 0. For angles greater than π/2 up to π, cos θ becomes negative, making the dot product negative. Therefore, a⃗ ⋅ b⃗ ≥ 0 only when 0 ≤ θ ≤ π/2.

If θ is the angle between two vectors a⃗ and b⃗, then a⃗ ⋅ b⃗ ≥ 0…

Question 65 of 68intermediate💡 UnderstandMCQ1 mark

If θ is the angle between two vectors a⃗ and b⃗, then a⃗ ⋅ b⃗ $\geq$ 0 only when

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 10 | Q: 16 | (Page 36)

For More Understanding

Explanation

Option B is correct because the dot product formula states that a⃗ ⋅ b⃗ = |a⃗||b⃗|cos θ. For the dot product to be non-negative ( $\geq$ 0), we need cos θ $\geq$ 0 since magnitudes |a⃗| and |b⃗| are always non-negative. The cosine function is non-negative when 0 $\leq$ θ $\leq$ π/2. At θ = 0, cos θ = 1, giving maximum positive value. At θ = π/2, cos θ = 0, making the dot product exactly 0. For angles greater than π/2 up to π, cos θ becomes negative, making the dot product negative. Therefore, a⃗ ⋅ b⃗ $\geq$ 0 only when 0 $\leq$ θ $\leq$ π/2.

Solution Steps

Step 1: Recall the dot product formula: a⃗ ⋅ b⃗ = |a⃗||b⃗|cos θ
Step 2: For a⃗ ⋅ b⃗ $\geq$ 0, we need |a⃗||b⃗|cos θ $\geq$ 0
Step 3: Since |a⃗| $\geq$ 0 and |b⃗| $\geq$ 0, the condition reduces to cos θ $\geq$ 0
Step 4: cos θ $\geq$ 0 when 0 $\leq$ θ $\leq$ π/2
Step 5: Therefore, the correct option is (B) 0 $\leq$ θ $\leq$ π/2

Previous Next